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THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 

GIFT  OF 

ALLEN  P.   SMITH 


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A  TREATISE 

ON 

ARCHITECTURE  AND  BUILDING 
CONSTRUCTION 

PREPARED  FOR  STUDENTS  OF 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS 

SCRANTON,  PA. 


Volume  I 


ARITHMETIC 
FORMULAS 

GEOMETRY    AND   MENSURATION 
ARCHITECTURAL  ENGINEERING 

WITH  PRACTICAL  QUESTIONS  AND  EXAMPLES 


First  Edition 


SCRANTON 

THE  COLLIERY  ENGINEER  CO. 
1899 


Entered  according  to  the  Act  of  Congress,  in  the  year  1899,  by  THE  COLLIERY 

ENGINEER  COMPANY,  in  the  office  of  the  Librarian  of 

Congress,  at  Washington. 


PRESS  OF  EATON  &  MAINS, 

NEW   YORK. 


l'i 
l/V 


PREFACE. 


In  the  first  six  volumes  of  the  eight  volumes  of  this  set 
are  comprised  all  the  Instruction  and  Question  Papers  used 
in  our  Complete  Architectural  Course ;  they  form  a  thorough, 
progressive,  and  comprehensive  treatise  on  the  subject  of 
Architecture. 

While  the  individual  Instruction  Papers  are  not  in  them- 
selves exhaustive  in  their  treatment  of  the  particular  subjects 
named  in  their  titles,  yet  they  are  so  closely  interrelated, 
one  with  another,  that  when  they  are  joined  together  in  one 
harmonious  whole  (as  in  these  volumes),  they  constitute  a 
treatise  that  is  complete  in  all  those  details  of  design  and 
construction  that  are  likely  to  be  met  with  in  general  archi- 
tectural practice.  The  student  can  therefore  use  them  as 
works  of  reference  in  connection  with  any  of  the  numerous 
problems  that  so  frequently  arise  in  all  branches  of  archi- 
tectural work. 

The  method  of  numbering  the  pages,  cuts,  articles,  etc. 
is  such  that  each  paper  and  part  is  complete  in  itself;  hence, 
in  order  to  make  the  indexes  intelligible,  it  was  necessary  to 
give  each  paper,  and  part  a  number.  This  number  is  placed 
at  the  top  of  each  page,  on  the  headline,  opposite  the  page 
number;  and  to  distinguish  it  from  the  page  number,  it  is 
preceded  by  the  printer's  section  mark  §.  Consequently,  a 
reference  such  as  Art.  29,  §  8,  would  be  readily  found  as 
follows:  The  back  stamp  on  each  volume,  except  Vols.  VII 


iv  PREFACE. 

and  VIII,  shows  the  sections  (i.e.,  papers)  included  in  the 
volume,  that  for  Vol.  II  reading  §§  7-10;  hence,  look  in 
Vol.  II  along  the  headlines  until  §  8  is  found,  and  then 
through  §  8  until  Art.  29  is  found. 

The  Question  Papers  are  given  the  same  section  numbers 
as  the  Instruction  Papers  to  which  they  belong,  and  are 
grouped  together  at  the  end  of  the  volumes  containing  the 
Instruction  Papers  to  which  they  refer.  The  paging  of  each 
Question  Paper  begins  with  (1),  as  in  the  case  of  the  Instruc- 
tion Papers. 

The  volumes  of  the  present  Course,  the  Complete  Archi- 
tectural, are  eight  in  number: 

Vol.  I  (§§  1-G)  contains  the  Instruction  and  Question 
Papers  on  Arithmetic,  Formulas,  Geometry  and  Mensuration, 
and  Architectural  Engineering. 

Vol.  II  (§§  7-10)  contains  the  Instruction  and  Question 
Papers  on  Masonry,  Carpentry,  and  Joinery. 

Vol.  Ill  (§§  11-15)  contains  the  Instruction  and  Question 
Pape'rs  on  Stair  Building,  Ornamental  Ironwork,  Roofing, 
Sheet-Metal  Work,  and  Electric-Light  Wiring  and  Bellwork. 

Vol.  IV  (§§  16-19)  contains  the  Instruction  and  Question 
Papers  on  Plumbing  and  Gas-Fitting,  Heating  and  Ventila- 
tion, Painting  and  Decorating,  and  Estimating  and  Calculating 
Quantities. 

Vol.  V  (§§  20-25)  contains  the  Instruction  and  Question 
Papers  on  History  of  Architecture,  Architectural  Design, 
Specifications,  Building  Superintendence,  and  Contracts  and 
Permits. 

Vol.  VI  contains  the  Drawing  Plates  and  the  instructions 
for  drawing  them.  Nothing  equal  to  this  volume  has  ever 
before  been  published.  It  forms  a  complete  course  in 
Architectural  Drawing.  For  convenience,  the  sections  are 
numbered  from  1  to  4,  instead  of  being  continued  from 
Vol.  V. 

Vol.  VII  contains  the  tables  and  formulas  given  in  the 
various  Instruction  Papers.  The  student  who  has  finished 
his  Course  will  find  this  volume  of  great  value.  All  the 
principal  formulas,  with  the  definitions  of  the  letters  used  in 


PREFACE.  v 

them,  are  conveniently  arranged  for  reference,  so  that  the 
student  can  save  the  labor  and  time  of  hunting  them  out  in 
the  Instruction  Papers. 

Vol.  VIII  contains  the  answers  to  the  questions  and  solu- 
tions to  the  examples  in  the  Question  Papers.  Whenever  it 
has  been  deemed  inadvisable  to  answer  a  question,  a  refer- 
ence to  the  proper  article  in  the  Instruction  Paper  has  been 
given,  the  reading  of  which  will  enable  the  student  to  answer 
the  question  himself. 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS. 


CONTENTS. 


ARITHMETIC.                                                            Section.  Page. 

Definitions        1  1 

Notation  and  Numeration 1  1 

Addition 1  4 

Subtraction 1  9 

Multiplication 1  11 

Division 1  16 

Cancelation 1  19 

Fractions 1  22 

Decimals 1  35 

Symbols  of  Aggregation 1  49 

Percentage 2  1 

Denominate  Numbers .  2  7 

Involution 2  22 

Evolution 2  25 

Ratio 2  42 

Proportion 2  46 

FORMULAS. 

Use  and  Application  of  Formulas   ...  3  1 

GEOMETRY  AND  MENSURATION. 

Lines  and  Angles 4  1 

Plane  Figures 4  6 

The  Triangle 4  7 

The  Circle 4  16 

Inscribed  and  Circumscribed  Polygons    .  4  21 

vii 


viii  CONTENTS. 

GEOMETRY  AND  MENSURATION — Continued.       Section.  Page. 

Mensuration 4  23 

Conversion  Tables 4  25 

Mensuration  of  Plane  Surfaces    ....  4  26 

Mensuration  of  Solids .  4  44 

Symmetrical  and  Similar  Figures    ...  4  54 

ARCHITECTURAL  ENGINEERING. 

Introduction 5  1 

The  Elements  of  Mechanics 5  3 

Definitions 5  3 

Effects  of  a  Force 5  3 

Composition  of  Forces 5  6 

Resolution  of  Forces 5  14 

Equilibrium 5  16 

Moments  of  Forces 5  17 

The  Lever 5  23 

Center  of  Gravity 5  26 

Loads  Carried  by  Structures 5  27 

Stresses  and  Strains 5  39 

Strength  of  Building  Materials   ....  5  42 

Foundations 5  48 

Columns 5  53 

Beams 5  62 

Reactions 5  62 

Stresses  in  Beams 5  69 

Strength  of  Beams 5  84 

Trussed  Beams 5  101 

Deflection  of  Floorbeams 5  110 

Graphical  Statics 5  111 

Design  for  a  Large  Building       ....  5  142 

Properties  of  Sections 6  1 

The  Neutral  Axis 6  1 

The  Moment  of  Inertia 6  8 

Resisting  Moment 6  14 

Radius  of  Gyration 6  15 

Steel  Columns 6  16 

Strength  of  Rivets  and  Pins  .  6  43 


CONTENTS.  ix 

ARCHITECTURAL  ENGINEERING — Continued.      Section.  Page. 

Plate  Girders G  57 

Deflection  of  Beams G  102 

Flitch  Plate  Girders G  10G 

Roof  Trusses G  110 

Determination    of    Stresses   in    the    Fink 

Truss G  110 

Design    of    a    Composite    Pin-Connected 

Roof  Truss G  118 

Design  of  a  Structural  Steel  Roof  Truss  .  6  124 
General  Notes  Regarding  the  Design  of  a 

Roof  Truss G  131 

Elements  of  Usual  Sections:  Table      .      .  G  135 

Value  of  Rivets :  Table G  130 

Areas  of  Angles:  Table G  137 

Properties  of  Angles,  Equal  Legs:  Table  G  138 
Properties    of    Angles,     Unequal    Legs: 

Table G  140 

Properties  of  Z  Bars:  Table G  141 

Properties  of  I  Beams:  Table     ....  G  143 

Properties  of  Channels:  Table    ....  G  144 

Radii  of  Gyration  for  Two  Angles :  Table  6  145 

Moduli  of  Elasticity:  Table 6  148 

Resisting  Moments  of  Pins:  Table       .      .  6  149 

Deflection  of  Beams:  Table G  152 

QUESTIONS  AND  EXAMPLES.  Section. 

Arithmetic , 1 

Arithmetic  (Continued) ,  2 

Formulas 3 

Geometry  and  Mensuration 4 

Architectural  Engineering 5 

Architectural  Engineering  (Continued)     ....  6 


ARITHMETIC. 


DEFIXITIOXS. 

1.  Arithmetic  is  the  art  of  reckoning,  or  the  study  of 
numbers. 

2.  A  unit  is  one,  or  a  single  thing,  as  one,  one  boy,  one 
horse,  one  dozen. 

3.  A  number  is  a  unit  or  a  collection  of  units,  as  one, 
three  apples,  five  boys. 

4.  The  unit  of  a  number  is  one  of  the  collection  of 
units  which  constitutes   the   number.      Thus,    the   unit   of 
twelve  is  one,  of  twenty  dollars  is  one  dollar. 

5.  A  concrete  number  is  a  number  applied  to  some 
particular  kind  of  object  or  quantity,  as  three  horses,  five 
dollars,  ten  pounds. 

6.  An  abstract  number  is  a  number  that  is  not  applied 
to  any  object  or  quantity,  as  three,  five,  ten. 

7.  Like  numbers  are  numbers  which  express  units  of 
the  same  kind,  as  six  days  and  ten  days,  two  feet  and  five  feet. 

8.  Unlike  numbers  are  numbers  which  express  units 
of  different  kinds,  as  ten  months  and  eight  miles,  seven  dol- 
lars and  five  feet. 

NOTATION  AND  NUMERATION. 

9.  Numbers  are  expressed  in  three  ways:    (1)  by  words; 
(2)  by  figures;  (3)  by  letters. 

10.  Notation  is  the  art  of  expressing  numbers  by  fig- 
ures or  letters. 

11.  Numeration   is  the  art  of  reading  the  numbers 
which  have  been  expressed  by  figures  or  letters. 


2  ARITHMETIC.  §  1 

12.  The  Arabic  notation  is  the  method  of  expressing 
numbers   by  figures.     This  method   employs  ten  different 
figures  to  represent  numbers,  viz. : 

Figures       0123456789 
Names  naught,  one    two  three  four  five     six  seven  eight  nine 

cipher, 

or  zero 

The  first  character  (0)  is  called  naught,  cipher,  or  zero, 
and  when  standing  alone  has  no  value. 

The  other  nine  figures  are  called  digits,  and  each  has  a 
value  of  its  own. 

Any  whole  number  is  called  an  integer. 

1 3.  As  there  are  only  ten  figures  used  in  expressing  num- 
bers, each  figure  must  have  a  different  value  at  different  times. 

14.  The  value  of  a  figure  depends  upon  its  position  in 
relation  to  others. 

15.  Figures   have  simple  values  and   local,  or  rela- 
tive, values. 

16.  The  simple  value  of  a  figure  is  the  value  it  ex- 
presses when  standing  alone. 

17.  The  local,  or  relative,  value  of  a  figure  is  the 
increased  value  it  expresses  by  having  other  figures  placed 
on  its  right. 

For  instance,   if  we  see  the  figure   6   standing 

alone,  thus 6 

we  consider  it  as  six  units,  or  simply  six. 

Place  another  6  to  the  left  of  it ;  thus 66 

The  original  figure  is  still  six  units,  but  the  sec- 
ond figure  is  ten  times  6,  or  6  tens. 

If  a  third  6  be  now  placed  still  one  place  further 
to  the  left,  it  is  increased  in  value  ten  times  more, 

thus  making  it  6  hundreds 666 

A  fourth  6  would  be  6  thousands 6666 

A  fifth   6  would  be  6   tens   of  thousands,   or 

sixty  thousands 66666 

A  sixth  6  would  be  6  hundreds  of  thousands  .     666666 
A  seventh  6  would  be  6  millions  .  .   6666666 


ARITHMETIC. 


The  entire  line  of  seven  figures  is  read  six  millions  six 
hundred  sixty-six  thousands  six  hundred  sixty-six. 

18.  The  increased  A'alue  of  each  of  these  figures  is  its 
local,  or  relative,  value.      Each  figure  is  ten  times  greater  in 
value  than  the  one  immediately  on  its  right. 

19.  The  cipher  (0)  has  no  value  in  itself,  but  it  is  useful 
in  determining  the  place  of  other  figures.     To  represent  the 
number  four  hundred  fire,   two  digits  only  are  necessary, 
one  to  represent  four  hundred,  and  the  other  to  represent 
five  units;  but  if  these  two  digits  are  placed  together,  as  -45, 
the  4  (being  in  the  second  place)  will  mean  4  tens.     To  mean 
4  hundreds,  the  4  should  have  two  figures  on  its  right,  and  a 
cipher  is  therefore  inserted  in  the  place  usually  given  to  tens, 
to  show  that  the  number  is  composed  of  hundreds  and  units 
only,  and  that  there  are  no  tens.     Four  hundred  five  is  there- 
fore expressed  as  405.      If  the  number  were  four  thousand 
and  five,  two  ciphers  would  be  inserted;  thus,  4005.      If  it 
were  four  hundred  fifty,  it  would  have  the  cipher  at  the 
right-hand  side  to  show  that  there  were  no  units,  and  only 
hundreds  and  tens ;    thus,    450.     Four  thousand  and  fifty 
would  be  expressed  4050,  the  first  cipher  indicating  that  there 
are  no  units  and  the  second  that  there  are  no  hundreds. 

20.  In  reading  numbers  that  have  been  represented  by 
figures,  it  is  usual  to  point  off  the  number  into  groups  of 
three  figures  each,  beginning  with  the  right-hand,  or  units, 
column,  a  comma  (,)  being  used  to  point  off  these  groups. 


Billions. 

Millions. 

Thousands. 

Units. 

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4  ARITHMETIC.  §  1 

In  pointing  off  these  figures,  begin  at  the  right-hand  figure 
and  count — units,  tens,  hundreds;  the  next  group  of  three 
figures  is  thousands;  therefore,  we  insert  a  comma  (,)  before 
beginning  with  them.  Beginning  at  the  figure  5,  we  say 
thousands,  tens  of  thousands,  hundreds  of  thousands,  and 
insert  another  comma.  We  next  read  millions,  tens  of  mil- 
lions, hundreds  of  millions  (insert  another  comma),  billions, 
tens  of  billions,  hundreds  of  billions. 

The  entire  line  of  figures  would  be  read:  four  hundred 
thirty-two  billions  one  hundred  ninety-eight  millions  seven 
hundred  sixty-Jive  thousands  four  hundred  thirty-two.  When 
we  thus  read  a  line  of  figures  it  is  called  numeration,  and 
if  the  numeration  be  changed  back  to  figures,  it  is  called 
notation. 

For  instance,  the  writing  of  the  following  figures, 

72,584,623, 

would  be  the  notation,  and  the  numeration  would  be  sev- 
enty-two millions  five  liundred  eighty-four  thousands  six  hun- 
dred twenty-three. 

21.  NOTE. — It  is  customary  to  leave  the  s  off  the  words  mil- 
lions, thousands,  etc.,  in  cases  like  the  above,  both  in  speaking  and 
writing;   hence,  the   above  would  usually  be  expressed  seventy -two 
million  five  hundred  eighty-four  thousand  six  hundred  twenty-three. 

22.  The  four  fundamental  processes  of  arithmetic  are 
addition,    subtraction,    multiplication,    and    division. 

They  are  called  fundamental  processes  because  all  operations 
in  arithmetic  are  based  upon  them. 


ADDITION. 

23.  Addition  is  the  process  of  finding  the  sum  of  two  or 
more  numbers.     The  sign  of  addition  is  +.     It  is  read//w^, 
and  means  more.      Thus,  5  +  6  is  read  5  plus  6,  and  means 
that  5  and  6  are  to  be  added. 

24.  The  sign   of  equality  is  = .     It  is  read  equals  or 
is   equal  to.      Thus,   5  +  6  =  11    may    be   read    5   plus   6 
equals  11. 


ARITHMETIC. 


25.  Like  numbers   can  be  added,  but  unlike  numbers 
cannot  be  added.    Thus,  0  dollars  can  be  added  to  7  dollars, 
and  the  sum  will  be  13  dollars;   but   G  dollars  cannot  be 
added  to  7  feet. 

26.  The  following  table  gives  the  sum  of  any  two  num- 
bers from  1  to  12: 


1  and    1  is    2 

2  and    1  is    3  ' 

3  and    1  is    4 

4  and    1  is    5 

1  and    2  is    3 

2  and    2  is    4 

3  and    2  is    5 

4  and    2  is    6 

1  and    3  is    4 

2  and    3  is    5 

3  and    3  is    6 

4  and    3  is    7 

1  and    4  is    5 

2  and    4  is    6 

3  and    4  is    7 

4  and    4  is    8 

1  and    5  is    6 

2  and    5  is    7 

3  and    5  is    8 

4  and    5  is    9 

1  and    6  is    7 

2  and    6  is    8 

3  and    6  is    9 

4  and    6  is  10 

1  and    7  is    8 

2  and    7  is    9 

3  and    7  is  10 

4  and    7  is  11 

1  and    8  is    9 

2  and    8  is  10 

3  and    8  is  11 

4  and    8  is  12 

1  and    9  is  10 

2  and    9  is  11 

3  and    9  is  12 

4  and    9  is  13 

1  and  10  is  11 

2  and  10  is  12 

3  and  10  is  13 

4  and  10  is  14 

1  and  11  is  12 

2  and  11  is  13 

3  and  11  is  14 

4  and  11  is  15 

1  and  12  is  13 

2  and  12  is  14 

3  and  12  is  15 

4  and  12  is  16 

5  and    1  is    6 

6  and    1  is    7 

7  and    1  is    8 

8  and    1  is    9 

5  and    2  is    7 

6  and    2  is    8 

7  and    2  is    9 

8  and    2  is  10 

5  and    3  is    8 

6  and    3  is    9 

7  and    3  is  10 

8  and    3  is  11 

5  and    4  is    9 

6  and    4  is  10 

7  and    4  is  11 

8  and    4  is  12 

5  and    5  is  10 

6  and    5  is  11 

7  and    5  is  12 

8  and    5  is  13 

5  and    6  is  11 

6  and    6  is  12 

7  and    6  is  13 

8  and    6  is  14 

5  and    7  is  12 

6  and    7  is  13 

7  and    7  is  14 

8  and    7  is  15 

5  and    8  is  13 

6  and    8  is  14 

7  and    8  is  15 

8  and    8  is  16 

5  and    9  is  14 

6  and    9  is  15 

7  and    9  is  16 

8  and    9  is  17 

5  and  10  is  15 

6  and  10  is  16 

7  and  10  is  17 

8  and  10  is  18 

5  and  11  is  16 

6  and  11  is  17 

7  and  11  is  18 

8  and  11  is  19 

5  and  12  is  17 

6  and  12  is  18 

7  and  12  is  19 

8  and  12  is  20 

9  and    1  is  10 

10  and    1  is  11 

11  and    1  is  12 

12  and    1  is  13 

9  and    2  is  11 

10  and    2  is  12 

11  and    2  is  13 

12  and    2  is  14 

9  and    3  is  12 

10  and    3  is  13 

11  and    3  is  14 

12  and    3  is  15 

9  and    4  is  13 

10  and    4  is  14 

11  and    4  is  15 

12  and    4  is  16 

9  and    5  is  14 

10  and    5  is  15 

11  and    5  is  16 

12  and    5  is  17 

9  and    6  is  15 

10  and    6  is  16 

11  and    6  is  17 

12  and    6  is  18 

9  and    7  is  16 

10  and    7  is  17 

11  and    7  is  18 

12  and    7  is  19 

9  and    8  is  17 

10  and    8  is  18 

11  and    8  is  19 

12  and    8  is  20 

9  and    9  is  18 

10  and    9  is  19 

11  and    9  is  20 

12  and    9  is  21 

9  and  10  is  19 

10  and  10  is  20 

11  and  10  is  21 

12  and  10  is  22 

9  and  11  is  20 

10  and  11  is  21 

11  and  11  is  22 

12  and  11  is  23 

9  and  12  is  21 

10  and  12  is  22 

11  and  12  is  23 

12  and  12  is  24 

This  table  should  be  carefully  committed  to  memory.  Since  0  has  no 
value,  the  sum  of  any  number  and  0  is  the  number  itself;  thus  17  and  0 
is  17. 

27.  For  addition,  place  the  numbers  to  be  added  directly 
under  each  other,  taking  care  to  place  units  under  units,  tens 
under  tens,  hundreds  under  hundreds,  and  so  on. 


6  ARITHMETIC.  §  1 

When  the  numbers  are  thus  written,  the  right-hand  figure 
of  one  number  is  placed  directly  under  the  right-hand  figure 
of  tJie  one  above  it,  thus  bringing  units  under  units,  tens 
under  tens,  etc.  Proceed  as  in  the  following  examples: 

28.  EXAMPLE.— What  is  the  sum  of  131,  222,  21,  2,  and  413  ? 

SOLUTION. —  131 

222 

21 

2 

413 
sum    789    Ans. 

EXPLANATION. — After  placing  the  numbers  in  proper 
order,  begin  at  the  bottom  of  the  right-hand,  or  units,  col- 
umn, and  add,  mentally  repeating  the  different  sums.  Thus, 
three  and  two  are  five  and  one  are  six  and  two  are  eight  and 
one  are  nine,  the  sum  of  the  numbers  in  units  column. 
Place  the  9  directly  beneath  as  the  first,  or  units,  figure  in 
the  stim. 

The  sum  of  the  numbers  in  the  next,  or  tens,  column 
equals  8  tens,  which  is  the  second,  or  tens,  figure  in  the 
sum. 

The  sum  of  the  numbers  in  the  next,  or  hundreds,  column 
equals  7  hundreds,  which  is  the  third,  or  hundreds,  figure  in 
the  sum. 

The  sum,  or  answer,  is  789. 

29.  EXAMPLE.— What  is  the  sum  of  425,  36,  9,215,  4,  and  907  ? 

SOLUTION. —  425 

36 

9215 
4 

907 
27 
60 

1  500 
9000 


sum    10587    Ans. 
EXPLANATION. — The  sum  of  the  numbers  in  the  first,  or 


§  1  ARITHMETIC.  7 

units,  column  is  seven  and  four  are  eleven  and  five  are  six- 
teen and  six  are  twenty-two  and  five  are  twenty-seven,  or 
27  units;  i.  e. ,  two  tens  and  seven  units.  Write  27  as 
shown.  The  sum  of  the  numbers  in  the  second,  or  tens, 
column  is  six  tens,  or  GO.  Write  00  underneath  27,  as 
shown.  The  sum  of  the  numbers  in  the  third,  or  hundreds, 
column  is  15  hundreds,  or  1,500.  WTrite  1,500  under  the  two 
preceding  results  as  shown.  There  is  only  one  number  in 
the  fourth,  or  thousands,  column,  9,  which  represents  9,000. 
Write  9,000  under  the  three  preceding  results.  Adding 
these  four  results,  the  sum  is  10,587,  which  is  the  sum  of 
425,  36,  9,215,  4,  and  907. 

NOTE. — It  frequently  happens  when  adding  a  long  column  of  fig- 
ures, that  the  sum  of  two  numbers,  one  of  which  does  not  occur  in  the 
addition  table,  is  required.  Thus,  in  the  first  column  above,  the  sum 
of  16  and  6  was  required.  We  know  from  the  table  that  6  +  6  —  12; 
hence,  the  first  figure  of  the  sum  is  2.  Now,  the  sum  of  any  number 
less  than  20  and  of  any  number  less  than  10  must  be  less  than  30,  since 
20  +  10  =  30;  therefore,  the  sum  is  22.  Consequently,  in  cases  of  this 
kind,  add  the  first  figure  of  the  larger  number  to  the  smaller  number, 
and  if  the  result  is  greater  than  9,  increase  the  second  figure  of  the  larger 
number  by  1.  Thus,  44  +  7  =  ?  4  +  7  =  11;  hence,  44  +  7  =  51. 

3O.     The  addition  may  also  be  performed  as  follows: 

425 
36 

9215 
4 

907 
sum  10587  Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  units  column 
is  27  units,  or  2  tens  and  7  units.  Write  the  7  units  as  the 
first,  or  right-hand,  figure  in  the  sum.  Reserve  the  two 
tens  and  add  them  to  the  figures  in  tens  column.  The  sum 
of  the  figures  in  the  tens  column,  plus  the  2  tens  reserved 
and  carried  from  the  units  column,  is  8,  which  is  written 
down  as  the  .second  figure  in  the  sum.  There  is  nothing  to 
carry  to  the  next  column,  because  -8  is  less  than  10.  The 
sum  of  the  numbers  in  the  next  column  is  15  hundreds,  or 
1  thousand  and  5  hundreds.  Write  down  the  5  as  the  third, 
or  hundreds,  figure  in  the  sum  and  carry  the  1  to  the  next 

1-2 


8  ARITHMETIC.  §  1 

column.     1  +  9  =  10,  which  is  written  down  at  the  left  of 
the  other  figures. 

The  second  method  saves  space  and  figures,  but  the  first 
is  to  be  preferred  when  adding  a  long  column. 

31.  EXAMPLE. — Add  the  numbers  in  the  column  below: 

SOLUTION.—  890 

82 

90 
393 
281 

80 
770 

83 
492 

80 
383 

84 
191 
sum  3899  Ans. 

EXPLANATION. — The  sum  of  the  digits  in  the  first  column 
equals  19  units,  or  1  ten  and  9  units.  Write  down  the  9  and 
carry  1  to  the  next  column.  The  sum  of  the  digits  in  the 
second  column  -f- 1  is  109  tens,  or  10  hundreds  and  9  tens. 
Write  down  the  9  and  carry  the  10  to  the  next  column. 
The  sum  of  the  digits  in  this  column  plus  the  10  reserved 
is  38. 

The  entire  sum  is  3,899. 

32.  Rule. — I.     Begin  at  the  right,  add  each  column  sep- 
arately, and  write  the  sum,  if  it  be  only  one  figure,  under  the 
column  added. 

II.  If  the  sum  of  any  column  consists  of  two  or  more 
figures,  put  the  right-hand  figure  of  the  sum  under  that 
column  and  add  the  remaining  figure  or  figures  to  the  next 
column. 

33.  Proof. —  To  prove  addition,  add  each  column  from 
top  to  bottom.     If  you  obtain  the  same  result  as  by  adding 
from  bottom  to  top,  the  work  is  probably  correct. 


1  ARITHMETIC. 

EXAMPLES    FOR    PRACTICE. 

34.  Find  the  sum  of: 

(a)  104  +  208  +  613  +  214.  \  (a)  1,134. 

(d)  1,875  +  3,143  +  5,826  +  10,832.  (6)  21,676. 

(c)  4,865  +  2,145  +  8,173  +  40,084.  (c)  55,267. 

(d)  14,204  +  8,173  +  1,065  +  10,042.  t    (d)  33,484. 

(e)  10,832  +  4,145  +  3,133  +  5,872.  ns"  |  (e)  23,982. 


(/)  214  +  1,231  +  141  +  5,000. 
(£•)  123  +  104  +  425  +  126  +  327. 


(/)     6,586. 
(-)     1,105. 


(h)     6,354  +  2,145  +  2,042  +  1,111  +  3,333.  [  (//)      14,985 


SUBTRACTION. 

35.  In  arithmetic,  subtraction  is  the  process  of  finding 
how  much  greater  one  number  is  than  another. 

The  greater  of  the  two  numbers  is  called  the  minuend. 
The  smaller  of  the  two  numbers  is  called  the  subtrahend. 
The  number  left  after  subtracting  the  subtrahend  from 
the  minuend  is  called  the  difference,  or  remainder. 

36.  The  sign  of    subtraction  is  —  .      It  is  read  minus, 
and  means  less.     Thus,  12  —  7  is  read  12  minus  7,  and  means 
that  7  is  to  be  taken  from  12. 

37.  EXAMPLE.— From  7,568  take  3,425. 
SOLUTION. —  minuend    7568 

subtrahend    3425 

remainder    4143     Ans. 

EXPLANATION. — Begin  at  the  right-hand,  or  units,  column 
and  subtract  in  succession  each  figure  in  the  subtrahend  from 
the  one  directly  above  it  in  the  minuend,  and  write  the  remain- 
ders below  the  line.  The  result  is  the  entire  remainder. 

38.  When  there  are  more  figures  in  the  minuend  than 
in  the  subtrahend,  and  when  some  figures  in  the  minuend 
are  less  than  the  figures  directly  under  them  in  the  subtra- 
hend, proceed  as  in  the  following  example : 

EXAMPLE.— From  8,453  take  844. 
SOLUTION. —  minuend    8453 

subtrahend       844 

remainder    7609    Ans. 


10  ARITHMETIC.  §  1 

EXPLANATION. — Begin  at  the  right-hand,  or  units,  column 
to  subtract.  We  cannot  take  4  from  3,  and  must,  therefore, 
borrow  1  from  5  in  tens  column  and  annex  it  to  the  3  in 
units  column.  The  1  ten  =  10  units,  which  added  to  the  3 
in  units  column  =  13  units.  4  from  13  =  9,  the  first,  or 
units,  figure  in  the  remainder. 

Since  we  borrowed  1  from  the  5,  only  4  remains ;  4  from 
4  =  0,  the  second,  or  tens,  figure.  We  cannot  take  8  from 
4,  and  must,  therefore,  borrow  1  from  8  in  thousands  colum'n. 
Since  1  thousand  =  10  hundreds,  10  hundreds  4- 4  hundreds 
=  14  hundreds,  and  8  from  14  =  6,  the  third,  or  hundreds, 
figure  in  the  remainder. 

Since  we  borrowed  1  from  8,  only  7  remains,  from  which 
there  is  nothing  to  subtract ;  therefore,  7  is  the  next  figure 
in  the  remainder,  or  answer. 

The  operation  of  borrowing  is  performed  by  mentally 
placing  1  before  the  figure  following  the  one  from  which  it 
is  borrowed.  In  the  above  example  the  1  borrowed  from  5 
is  placed  before  3,  making  it  13,  from  which  we  subtract  4. 
The  1  borrowed  from  8  is  placed  before  4,  making  14,  from 
which  8  is  taken. 

39.  EXAMPLE.— Find  the  difference  between  10,000  and  8,763. 

SOLUTION. —  minuend  10000 
subtrahend  8763 
remainder  1237  Ans. 

EXPLANATION. — In  the  above  example  we  borrow  1  from 
the  second  column  and  place  it  before  0,  making  10;  3 
from  10  =  7.  In  the  same  way  we  borrow  1  and  place  it 
before  the  next  cipher,  making  10 ;  but  as  we  have  borrowed 
1  from  this  column  and  have  taken  it  to  the  units  column, 
only  9  remains  from  which  to  subtract  6 ;  6  from  9  =  3. 
For  the  same  reason  we  subtract  7  from  9  and  8  from  9  for 
the  next  two  figures,  and  obtain  a  total  remainder  of  1,237. 

40.  Rule. — Place   the  subtrahend  (or  smaller  number] 
under  the  minuend  (or  larger  number},  in  the  same  manner  as 
for  addition,  and  proceed  as  in  Arts.  37,  38,  and  39. 


ARITHMETIC. 


11 


41.  Proof. —  To  prove  an  example  in  subtraction,  add 
tJie  subtrahend  and  the  remainder.  The  sum  should  equal 
the  minuend.  If  it  does  not,  a  mistake  has  been  made,  and 
the  work  should  be  done  over. 

Proof  of  the  above  example : 

subtrahend       8763 

remainder       1237 

minuend    10000 


42. 

(a) 

(d) 

(f) 
(h) 


EXAMPLES  FOR  PRACTICE. 

From: 

94,278  take  62,574. 
53,714  take  25,824. 
71,832  take  58,109. 
20,804  take  10,408. 
310,465  take  102,141. 
(81,043  +  1,041)  take  14,831. 
(20,482  +  18,216)  take  21,214. 


Ans. 


(2,040  H-  1,213  +  542)  take  3,791. 


31,704. 

27,890. 

13,723. 

10,396. 

208,324. 

67,253. 

17,484. 


(//)      4. 


MULTIPLICATION. 

43.  To  multiply  a  number  is  to  add  it  to  itself  a  certain 
number  of  times. 

44.  Multiplication  is  the  process  of  multiplying'  one 
number  by  another. 

The  number  thus  added  to  itself,  or  the  number  to  be 
multiplied,  is  called  the  multiplicand. 

The  number  which  shows  how  many  times  the  multi- 
plicand is  to  be  taken,  or  the  number  by  which  we  multiply, 
is  called  the  multiplier. 

The  result  obtained  by  multiplying  is  called  the  product. 

45.  The  sign  of  multiplication  is  X .      It  is  read  times  or 
multiplied  by.     Thus,  9  X  6  is  read  9  times  6,  or  9  multiplied 
by  6. 

46.  It  matters  not  in  what  order  the  numbers  to  be  multi- 
plied together  are  placed.     Thus,  G  X  9  is  the  same  as  9  X  C. 


12 


ARITHMETIC. 


1 


47.     In  the  following  table,  the  product  of  any  two  num- 
bers (neither  of  which  exceeds  12)  may  be  found: 


1  times    1  is      1 

2  times    1  is      2 

3  times    1  is      3 

1  times    2  is      2 

2  times    2  is      4 

3  times    2  is      6 

1  times    3  is      3 

2  times    3  is      6 

3  times    3  is      9 

1  times    4  is      4 

2  times    4  is      8 

3  times    4  is    12 

1  times    5  is      5 

2  times    5  is    10 

3  times    5  is    15 

1  times    6  is      6 

2  times    6  is    12 

3  times    6  is    18 

1  times    7  is      7 

2  times    7  is    14 

3  times    7  is    21 

1  times    8  is      8 

2  times    8  is    16 

3  times    8  is    24 

1  times    9  is      9 

2  times    9  is    18 

3  times    9  is    27 

1  times  10  is    10 

2  times  10  is    20 

3  times  10  is    30 

1  times  11  is    11 

2  times  11  is    22 

3  times  11  is    33 

1  times  12  is    12 

2  times  12  is    24 

3  times  12  is    36 

4  times    1  is      4 

5  times    1  is      5 

6  times    1  is      6 

4  times    2  is      8 

5  times    2  is    10 

6  times    2  is    12 

4'times    3  is    12 

5  times    3  is    15 

6  times    3  is    18 

4  times    4  is    16 

5  times    4  is    20 

6  times    4  is    24 

4  times    5  is    20 

5  times    5  is    25 

6  times    5  is    30 

4  times    6  is    24 

5  times    6  is    30 

6  times    6  is    36 

4  times    7  is    28 

5  times    7  is    35 

6  times    7  is    42 

4  times    Sis    32 

5  times    8  is    40 

6  times    8  is    48 

4  times    9  is    36 

5  times    9  is    45 

6  times    9  is    54 

44times  10  is    40 

5  times  10  is    50 

6  times  10  is    60 

4  times  11  is    44 

5  times  11  is    55 

6  times  11  is    66 

4  times  12  is    48 

5  times  12  is    60 

6  times  12  is    72 

7  times    1  is      7 

8  times    1  is      8 

9  times    1  is      9 

7  times    2  is    14 

8  times    2  is    16 

9  times    2  is    18 

7  times    3  is    21 

8  times    3  is    24 

9  times    3  is    27 

7  times    4  is    28 

8  times    4  is    32 

9  times    4  is    36 

7  times    5  is    35 

8  times    5  is    40 

9  times    5  is    45 

7  times    6  is    42 

8  times    6  is    48 

9  times    6  is    54 

7  times    7  is    49 

8  times    7  is    56 

9  times    7  is    63 

7  times    8  is    56 

8  times    8  is    64 

9  times    8  is    72 

7  times    9  is    63 

8  times    9  is    72 

9  times    9  is    81 

7  times  10  is    70 

8  times  10  is    80 

9  times  10  is    90 

7  times  11  is    77 

8  times  11  is    88 

9  times  11  is    99 

7  times  12  is    84 

8  times  12  is    96 

9  times  12  is  108 

10  times    1  is    10 

11  times    1  is    11 

12  times    1  is    12 

10  times    2  is    20 

11  times    2  is    22 

12  times    2  is    24 

10  times    3  is    30 

11  times    3  is    33 

12  times    3  is    36 

10  times    4  is    40 

11  times    4  is    44 

12  times    4  is    48 

10  times    5  is    50 

11  times    5  is    55 

12  times    5  is    60 

10  times    6  is    60 

11  times    6  is    66 

12  times    6  is    72 

10  times    7  is    70 

11  times    7  is    77 

12  times    7  is    84 

10  times    8  is    80 

11  times    8  is    88 

12  times    Sis    96 

10  times    9  is    90 

11  times    9  is    99 

12  times    9  is  108 

10  times  10  is  100 

11  times  10  is  110 

12  times  10  is  120 

10  times  11  is  110 

11  times  11  is  121 

12  times  11  is  132 

10  times  12  is  120 

11  times  12  is  133 

12  times  12  is  144 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  product  of  0  and  any  number  is  0. 


§  1  ARITHMETIC.  13 

48.     To  multiply  a  number  by  one  figure  only: 

EXAMPLE. — Multiply  425  by  5. 
SOLUTION. —        multiplicand        425 
multiplier  5 

product      2125     Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  right-hand  figure  of  the  multiplicand. 
On  looking  in  the  multiplication  table,  we  see  that  5x5  are 
25.  Multiplying  the  first  figure  at  the  right  of  the  multi- 
plicand, or  5,  by  the  multiplier,  5,  it  is  seen  that  5  times  5 
units  are  25  units,  or  2  tens  and  5  units.  Write  the  5  units 
in  units  place  in  the  product,  and  reserve  the  2  tens  to  add 
to  the  product  of  tens.  Looking  in  the  multiplication  table 
again,  we  see  that  5x2  are  10.  Multiplying  the  second 
figure  of  the  multiplicand  by  the  multiplier,  5,  we  see  that 
5  times  2  tens  aYe  10  tens,  and  10  tens  plus  the  2  tens  reserved 
are  12  tens,  or  1  hundred  plus  2  tens.  Write  the  2  tens  in 
tens  place,  and  reserve  the  1  hundred  to  add  to  the  product 
of  hundreds.  Again,  we  see  by  the  multiplication  table  that 
5x4  are  20.  Multiplying  the  third,  or  last,  figure  of  the 
multiplicand  by  the  multiplier,  5,  we  see  that  5  times  4  hun- 
dreds are  20  hundreds,  and  20  hundreds  plus  the  1  hundred  re- 
served are  21  hundreds,  or  2  thousands  and  1  hundred,  which 
we  write  in  thousands  and  hundreds  places,  respectively. 

Hence,  the  product  is  2,125. 

This  result  is  the  same  as  adding  425  five  times.     Thus, 

425 
425 
425 
425 
425 
sum  2125  Ans. 


EXAMPLES  FOR  PRACTICE. 

49.      Find  the  product  of  : 


(a)  61,483  X  6. 

(b)  12,375  X  5. 

(c)  10,426  X  7. 

(d)  10,835  X  3. 


Ans.  * 


(a)  368,898. 

(V)  61,875. 

(c)  72,982. 

(<t)  32,505. 


14  ARITHMETIC.  §  1 

(e)  98,376X4.  f  (e)  393,504. 

(/)  10,873X8.  (/)  86,984. 

(g)  71,543X9.  "   (£•)  643,887. 

(h)  218,734x2.  [(*)  437,468. 

50.  To  multiply  a  number  by  two  or  more  figures : 

EXAMPLE.— Multiply  475  by  234. 

SOLUTION. —     multiplicand  475 

multiplier  234 

1900 
1425 
950 
product    111150    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  multiplicand,  placing  units  under 
units,  tens  under  tens,  etc. 

We  cannot  multiply  by  234  at  one  operation;  we  must, 
therefore,  multiply  by  the  parts  and  then  add  the  partial 
products. 

The  parts  by  which  we  are  to  multiply  are  4  units,  3  tens, 
and  2  hundreds.  4  times  475  =  1,900,  the  first  partial 
product ;  3  times  475  =  1,425,  the  second  partial  product, 
the  right-hand  figure  of  which  is  written  directly  under 
the  figure  multiplied  by,  or  3 ;  2  times  475  =  950,  the  tliird 
partial  product,  the  right-hand  figure  of  which  is  written 
directly  under  the  figure  multiplied  by,  or  2. 

The  sum  of  these  three  partial  products  is  111,150,  which 
is  the  entire  product. 

51.  Rule. — I.      Write  the  multiplier  under  the  multipli- 
cand, so  that  units  are  under  units,  tens  under  tens,  etc. 

II.  Begin  at  the  right  and  multiply  each  figure  of  the 
multiplicand  by  each    successive  figure   of  the   multiplier, 
placing  the  right-hand  figure  of  each  partial  product  directly 
under  the  figure  used  as  a  multiplier. 

III.  The  sum   of  the  partial  products   will  equal   the 
required  product. 


ARITHMETIC. 


15 


52.  Proof. — Review  tJic  work  carefully,  or  multiply  the 
multiplier  by  the  multiplicand ;  if  the  results  agree,  the  work 
is  correct. 

53.  When  there  is  a  cipher  in  the  multiplier,  multiply 
by  it  the  same  as  with  the  other  figures.     Thus, 

(a)  (6)  (c)  (d) 

0  2  15  708 

X_0_  X  0  X      0  X         0 

0    Ans.         0    Ans.  0  0    Ans.         000    An*. 


3114 
203 
9342 
0000 
6228 
632142  Ans. 


4008 
305 


20040 

0000 
1  2034 
1222440  Ans. 


31264 
1002 
62528 
00000 
00000 
31264 


31326528  Ans. 


When  multiplying  by  a  number  containing  a  cipher,  the 
work  may  be  shortened  by  writing  the  first  cipher  of  the 
partial  product,  then  multiplying  by  the  next  figure  of 
the  multiplier  and  writing  the  partial  product  alongside  of 
the  cipher.  Thus,  examples  (e]  and  (g)  above  might  have 
been  solved  in  the  following  manner : 


3114 
203 

9342 
62280 
632142  Ans. 


31264 
1002 

62528 
3126400 
31326528  Ans. 


54. 

(«) 
0) 
W 

Oo 
(*) 


EXAMPLES    FOR    PRACTICE. 

Find  the  product  of: 


3,842  X  26. 
3,716  X  45. 
1,817  X  134. 
675  X  38. 
1,875  X  33. 
4,836  X  47. 
5,682  X  543. 
3,257  X  246. 


Ans. 


r  («) 


(0 


(g) 


99,892. 

167,220. 

225,308. 

25,650. 

61,875. 

227,292. 

3,085,326. 

801,222. 


16  ARITHMETIC.  §  1 


(*)  2,875  X  302. 

(/)  17,819  X  1,004. 

(Jt)  38,674  X  205. 

(/)  18,304X100.  Ans. 

(M)  7,834  X  10. 

(»)  87,543X1,000. 

(o)  48,763  X  100. 


(z)  868,250. 

(j)  17,890,376. 

(k)  7,928,170. 

(/)  1,830,400. 

(m)  78,340 

(«)  87,543,000. 

I  (<?)  4,876,300. 


DIVISION. 

55.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another  of  the  same  kind. 

The  number  to  be  divided  is  called  the  dividend. 
The  number  by  which  we  divide  is  called  the  divisor. 
The  number  which  shows  how  many  times  the  divisor  is 
contained  in  the  dividend  is  called  the  quotient. 

56.  The  sign  of  division  is  -K      It  is  read  divided  by. 
54-7-9  is  read  54  divided  by  9.      Another  way  to  write  54 

54  54 

divided  by  9  is  -y.      Thus,  54  -j-  9  =  6,  or  y  =  6. 

In  both  of  these  cases,  54  is  the  dividend  and  9  is  the 
divisor. 

Division  is  the  reverse  of  multiplication. 

57.  To  divide  -when  the  divisor  consists  of  but  one 
figure,  proceed  as  in  the  following  example : 

EXAMPLE. — What  is  the  quotient  of  875  -=-  7  ? 

divisor  dividend  quotient 
SOLUTION.—  7)875(125    Ans. 

_7_ 

17 

14 


35 
35 

remainder        0 


EXPLANATION.—  7  is  contained  in  8  hundreds,  1  hundred 
times.  Place  the  1  as  the  first,  or  left-hand,  figure  of  the 
quotient.  Multiply  the  divisor,  7,  by  the  1  hundred  of  the 


§  1  ARITHMETIC.  17 

quotient,  and  place  the  product,  7  hundreds,  under  the  8 
hundreds  in  the  dividend,  and  subtract.  Beside  the  re- 
mainder, 1,  bring  down  the  next,  or  tens,  figure  of  the 
dividend,  in  this  case  7,  making  IT  tens;  7  is  contained  in 
17,  2  times.  Write  the  2  as  the  second  figure  of  the  quo- 
tient. Multiply  the  divisor,  7,  by  the  2  in  the  quotient,  and 
subtract  the  product  from  IT.  Beside  the  remainder,  3, 
bring  down  the  units  figure  of  the  dividend,  making  35  units. 
7  is  contained  in  35,  5  times,  which  is  placed  in  the  quotient. 
Multiplying  the  divisor  by  the  last  figure  of  the  quotient, 
5  times  7  =  35,  which  subtracted  from  35,  under  which  it 
is  placed,  leaves  0.  Therefore,  the  quotient  is  125.  This 
method  is  called  long  division. 

58.  In  short  division,  only  the  divisor,  dividend,  and 
quotient  are  written,  the  operations  being  performed  men- 
tally. 

dividend 

divisor  7  )  8  l  7 3  5 

quotient         125    Ans. 

The  mental  operation  is  as  follows:  7  is  contained  in  8, 
once  and  1  remainder;  imagine  1  to  be  placed  before  7, 
making  17;  7  is  contained  in  17,  2  times  and  3  over;  imag- 
ine 3  to  be  placed  before  5,  making  35 ;  7  is  contained  in  35, 
5  times.  These  partial  quotients,  placed  in  order  as  they 
are  found,  make  the  entire  quotient  125. 

59.  If  the  divisor  consists  of  two  or  more  figures,  pro- 
ceed as  in  the  following  example : 

EXAMPLE.— Divide  2,702,826  by  63. 

divisor         dividend  quotient 

SOLUTION.—  63)2702826(42902    Ans. 

252 
182 
126 


568 

567 
126 
126 


18  ARITHMETIC.  §  1 

EXPLANATION. — As  63  is  not  contained  in  the  first  two 
figures,  27,  we  must  use  the  first  three  figures,  270.  Now, 
by  trial  we  must  find  how  many  times  63  is  contained  in  270. 
6  is  contained  in  the  first  two  figures  of  270,  4  times.  Place  the 
4  as  the  first,  or  left-hand,  figure  in  the  quotient.  Multiply  the 
divisor,  63,  by  4,  and  subtract  the  product,  252,  from  270. 
The  remainder  is  18,  beside  which  we  write  the  next  figure 
of  the  dividend,  2,  making  182.  Now,  6  is  contained  in  the 
first  two  figures  of  182,  3  times,  but  on  multiplying  63  by  3, 
we  see  that  the  product,  189,  is  too  great,  so  we  try  2  as 
the  second  figure  of  the  quotient.  Multiplying  the  divi- 
sor, 63,  by  2  and  subtracting  the  product,  126,  from  182,  the 
remainder  is  56,  beside  which  we  bring  down  the  next  figure 
'of  the  dividend,  making  568.  6  is  contained  in  56  about  9 
times.  Multiply  the  divisor,  63,  by  9  and  subtract  the  prod- 
uct, 567,  from  568.  The  remainder  is  1,  and  bringing 
down  the  next  figure  of  the  dividend,  2,  gives  12.  As  12  is 
smaller  than  63,  we  write  0  in  the  quotient  and  bring  down 
the  next  figure,  6,  making  126.  63  is  contained  in  126,  2 
times,  without  a  remainder.  Therefore,  42,902  is  the 
quotient. 

6O.  Rule. — I.  Write  the  divisor  at  the  left  of  the  divi- 
dend, with  a  line  between  them. 

II.  Find  how  many  times  the  divisor  is  contained  in  the 
lowest  number  of  the  left-hand  figures  of  the  dividend  that 
will  contain  it,  and  write  the  result  at   the  right  of  the 
dividend,   with  a  line  between,  for  the  first  figure  of  the 
quotient. 

III.  Multiply  the  divisor  by  this  quotient;  write  the  prod- 
uct under  the  partial  dividend  used,  and  subtract,  annexing 
to  the  remainder  the  next  figure  of  the  dividend.     Divide  as 
before,  and  thus  continue  until  all  tlie  figures  of  the  dividend 
have  been  used. 

IV.  If  any  partial  dividend  will  not  contain  the  divisor, 
write  a  cipher  in  the  quotient,  annex  the  next  figure  of  the 
dividend,  and  proceed  as  before. 


ARITHMETIC. 


19 


V.  If  there  be  at  last  a  remainder,  wife  it  after  the 
quotient,  with  the  divisor  underneath. 

61.  Proof. — Multiply  the  quotient  by  the  divisor  and  add 
the  remainder,  if  there  be  any,  to  the  prodnet.  The  result 
will  be  the  dividend.  Thus, 

divisor    dividend    quotient 
63)4235(67  J|     Ans. 


PROOF.- 


remainder 

quotient 
divisor 


remainder 
dividend 


4  5  5 

44  1 

1  4 

6  7 

63 

2  0  1 

402 

4221 

1  4 

423  5 


EXAMPLES   FOR   PRACTICE. 

62. 

Divide  the  following: 

(a) 

126,498  by  58.                                                  f  («) 

2,181. 

(b) 

3,207,594  by  767. 

(6) 

4,182. 

(c) 

11,408,202  by  234. 

(<•) 

48,753. 

(d) 

2,100,315  by  581. 

Ans.  • 

(>!} 

3,615. 

(') 

969,936  by  4,008. 

(<') 

242 

(/) 

7,481,888  by  1,021. 

(/) 

7,328. 

(£") 

1,525,915  by  5,003. 

(A'") 

305. 

(A) 

1,646,301  by  381. 

(/<) 

4,321. 

CAXCELATIOX. 

63.  Cancelatlou   is  the  process  of  shortening  opera- 
tions in  division  by  casting  out  equal   factors  from   both 
dividend  and  divisor. 

64.  The  factors  of  a  number  are  those  numbers,  which, 
when  multiplied  together,  produce  the  given  number.    Thus, 
5  and  3  are  the  factors  of  15,  since  5  X '5  --  15.      Likewise, 
8  and  7  are  the  factors  of  5G,  since 8X 7  =  5G. 


20  ARITHMETIC.  §  1 

65.  A  prime  number  is  one  which  cannot  be  divided 
by  any  number  except  itself  and  1.     Thus,  2,  3,  11,  29,  etc. 
are  prime  numbers. 

66.  A   prime   factor   is    any  factor   that   is    a   prime 
number. 

Any  number  that  is  not  a  prime  is  called  a  composite 
number,  and  may  be  produced  by  multiplying  together  its 
prime  factors.  Thus,  60  is  a  composite  number,  and  is 
equal  to  the  product  of  its  prime  factors,  2x2x3x5. 

67.  Canceling  equal  factors  from   both   dividend  and 
divisor  does  not  change  the  quotient. 

The  canceling  of  a  factor  in  both  dividend  and  divisor  is 
the  same  as  dividing  them  both  by  the  same  number,  and 
this,  evidently,  does  not  change  the  quotient. 

Write  the  numbers  forming  the  dividend  above  a  hori- 
zontal line,  and  those  forming  the  divisor  below  it;  then 
cancel  the  equal  factors. 

68:      EXAMPLE.—  Divide  4  X  45  X  60  by  9  X  24- 

SOLUTION.  —  Placing  the  dividend  over  the  divisor,  and  canceling, 

5       10 
*XffXPP_60  A 

PX#     -T- 

0 

EXPLANATION.  —  The  4  in  the  dividend  and  the  24  in  the 
divisor  are  both  divisible  by  4,  since  4  divided  by  4  equals  1, 
and  24  divided  by  4  equals  6.  Cross  off  the  4  and  write  the 
1  over  it  ;  also,  cross  off  the  24  and  write  the  6  under  it.  Thus, 

1 


6 

60  in  the  dividend  and  6  in  the  divisor  are  divisible  by  6, 
since  60  divided  by  6  equals  10,  and  6  divided  by  6  equals  1. 
Cross  off  the  60  and  write  10  over  it;  also,  cross  off  the  6 
and  write  1  under  it.  Thus, 

1  10 

j  X  45  X  99  _ 


ARITHMETIC. 


Again,  45  in  the  dividend  and  9  in  the  divisor  are  divisi- 
ble by  9,  since  45  divided  by  9  equals  5,  and  9  divided  by  9 
equals  1.  Cross  off  the  45  and  write  the  5  over  it;  also, 
cross  off  the  9  and  write  the  1  under  it.  Thus, 


10 

X  W 


Since  there  are  no  two  remaining  numbers  (one  in  the 
dividend  and  one  in  the  divisor)  divisible  by  any  number 
except  1,  without  a  remainder,  it  is  impossible  to  cancel 
further. 

Multiply  all  the  uncanceled  numbers  in  the  dividend 
together  and  divide  their  product  by  the  product  of  all  the 
uncanceled  mimbers  in  the  divisor.  The  result  will  be  the 
quotient.  The  product  of  all  the  uncanceled  numbers  in 
the  dividend  equals  5  X  1  X  10  =  50;  the  product  of  all  the 
uncanceled  numbers  in  the  divisor  equals  1  X  1  =  1. 


Hence-  =  =  5a  Ans- 


69.  Rule.  —  I.  Cancel  the  common  factors  from  both  the 
dividend  and  the  divisor. 

II.  Then  divide  the  product  of  the  remaining  factors  of  the 
dividend  by  the  product  of  the  remaining'  factors  of  the 
divisor,  and  the  result  will  be  the  quotient. 


EXAMPLES    FOR    PRACTICE. 
7O.      Divide: 

(a)  14X18X16X40  by  7X8X6X5X3. 

(b)  3  X  65  X  50  X  100  X  60  by  30  X  60  X  13  X  10. 

(c)  8X4X.3X9XH  by  11X9X4X3X8. 

(<t)     164X321X6X7X4  by  82X321X7-  . 

(e)      50  X  100X200X72  by  1,000X144X100. 
(/)    48  X  63X55X49  by  7  X  21  X  H  X  48. 


(g)    110  X  150  X  84  X  32  by  11X15  X  100X64. 

(h)     115  X  120X400  XI, 000  by  23X1,000X60X800.          [  (//)     5. 


(a)  82. 

(b)  250. 

(0  I- 

(if)  4K. 

(e)  5. 

(/)  105. 


22  ARITHMETIC. 

FRACTIONS. 


DEFINITIONS. 

71.  A  fraction   is  a  part  of  a   unit.      One-half,   one- 
third,  two-fiftlis  are  fractions. 

72.  Two  numbers  are  required  to  express  a  fraction ;  one 
is  called  the  numerator,  and  the  other,  the  denominator. 

73.  The  numerator  is  placed  above  the  denominator, 
with  a  line  between  them,   as  |.      Here,   3  is  the  denom- 
inator, and  shows  into  how  many  equal  parts  the  unit,  or 
one,  is  divided.     The  numerator,    2,    shows  how  many  of 
these  equal  parts  are  taken  or  considered.     The  denomi- 
nator also  indicates  the  names  of  the  parts. 

\  is  read  one-half. 

f  is  read  three-fourths. 

f  is  read  three-eighths. 
Y5¥  is  read  five-sixteenths, 
f-f-  is  read  twenty-nine  forty-sevenths. 

74.  In  the  expression  "  f  of  an  apple,"  the  denominator, 
4,  shows  that  the  apple  is  to  be  (or  has  been)  cut  into  4  equal 
parts,  and  the  numerator,  3,  shows  that  three  of  these  parts, 
or  fourths,  are  taken  or  considered. 

If  each  of  the  parts,  or  fourths,  of  the  apple  were  cut 
into  two  equal  pieces,  there  would  then  be  twice  as  many 
pieces  as  before,  or  4x2  =  8  pieces  in  all,  one  of  these 
pieces  would  be  called  one-eighth,  and  would  be  expressed 
in  figures  as  ^.  Three  of  these  pieces  would  be  called 
three-eighths,  and  written  f.  The  words  three-fourths, 
three-eighths,  five-sixteenths,,  etc.  are  abbreviations  of  three 
one-fourths,  three  one-eighths,  five  one-sixteenths,  etc.  It 
is  evident  that  the  larger  the  denominator,  the  greater  is  the 
number  of  parts  into  which  anything  is  divided;  conse- 
quently, the  parts  themselves  are  smaller,  and  the  value  of 
the  fraction  is  less  for  the  same  number  of  parts  taken.  In 
other  words,  £,  for  example,  is  smaller  than  £,  because  if  an 
object  be  divided  into  9  parts,  the  parts  are  smaller  than  if 


§  1  ARITHMETIC.  23 

the  same  object  had  been  divided  into  8  parts;  and,  since  ^ 
is  smaller  than  -*-,  it  is  clear  that  7  one-ninths  is  a  smaller 
amount  than  7  one-eighths.  Hence,  also,  f  is  less  than  f. 

75.  The  value  of  a  fraction  is  the  numerator  divided 
by  the  denominator,  as  |-  =  2,  f  =  '3. 

76.  The  line  between  the  numerator  and  the  denomi- 
nator means  divided  by,  or  -h . 

f  is  equivalent  to  3  -=-  4. 
•f-  is  equivalent  to  5  -f-  8. 

77.  The  numerator  and  the  denominator  of  a  fraction 
are  called  the  terms  of  a  fraction. 

78.  The  value  of  a  fraction  whose  numerator  and  denom- 
inator are  equal  is  1. 

\,  or  four-fourths  —  1. 
f-,  or  eight-eighths  =  1. 
ff,  or  sixty-four  sixty-fourths  =  1. 

79.  A  propei*  fraction  is  a  fraction  whose  numerator 
is  less  than  its  denominator.      Its  value  is  less  than  1,   as 

I.  i  *• 

80.  An  Improper  fraction  is  a  fraction  whose  numer- 
ator equals  or  is  greater  than  the  denominator.      Its  value  is 
1  or  more  than  1,  as  f ,  |,  f|. 

81.  A  mixed  number  is  a  whole  number  and  a  fraction 
united.     4f  is  a  mixed  number,  and  is  equivalent  to  4  -4- 1. 
It  is  read  four  and  two-tliirds. 


REDUCTION   OF  FRACTIONS. 

82.  Reduction  of  fractions  is  the  process  of  changing 
their  form  without  changing  their  value. 

83.  A  fraction  is  reduced  to  higher  terms  by  multiplying 
both  terms  of  the  fraction  by  the  same  number.     Thus,  J  is 
reduced  to  |  by  multiplying  both  terms  by  2. 

3X2  _    6 
4X2  "  8' 

1-3 


24  ARITHMETIC.  §  1 

The  value  is  not  changed,  since  J-  =  |.  For,  suppose 
that  an  object,  say  an  apple,  is  divided  into  8  equal  parts. 
If  these  parts  be  arranged  into  4  piles,  each  containing  2 
parts,  it  is  evident  each  pile  will  be  composed  of  the  same 
amount  of  the  entire  apple  as  would  have  been  the  case  had 
the  apple  been  originally  cut  into  4  equal  parts.  Now,  if 
one  of  these  piles  (containing  2  parts)  be  removed,  there  will 
be  3  piles  left,  each  containing  2  equal  parts,  or  6  equal  parts 
in  all,  i.  e. ,  six-eighths.  But,  since  one  pile,  or  one-quarter, 
was  removed,  there  are  three-quarters  left.  Hence,  f  —  f . 
The  same  course  of  reasoning  may  be  applied  to  any  similar 
case.  Therefore,  multiplying  both  terms  of  a  fraction  by 
the  same  number  does  not  alter  its  value. 

84.  To  reduce  a  fraction  to  an  equal  fraction  hav- 
ing a  given  denominator : 

EXAMPLE. — Reduce  %  to  an  equal  fraction  having  96  for  a  denominator. 

SOLUTION. — Both  the  numerator  and  the  denominator  must  be  mul- 
tiplied -by  the  same  number  in  order  not  to  change  the  value  of 
the  fraction.  The  denominator  must  be  multiplied  by  some  number 
which  will,  in  this  case,  make  the  product  96 ;  this  number  is  evidently 

7  v  12        84 

96  H-  8  =  12,  since  8  X  12  =  96.     Hence,  3.*  =  ^.     Ans. 

o  X  1*       «"* 

85.  Rule. — Divide  the  given  denominator  by  the  denom- 
inator of  the  given  fraction,  and  multiply  both  terms  of  the 
fraction  by  the  result. 

EXAMPLE. — Reduce  f  to  lOOths. 

SOLUTION.—    100  H- 4  =  25;  hence,  ?  X  ^  =  SL     Ans. 

4  X  25        100 

86.  A  fraction  is  reduced  to  lo^uer  terms  by  dividing 
both  terms  by  the  same  number.     Thus,  -fa  is  reduced  to  ^ 
by  dividing  both  terms  by  2. 

_8_-=-2  __  4 
10-i-2  "  5' 

That  -j^-  =  f  is  readily  seen  from  the  explanation  given 
in  Art.  83 ;  for,  multiplying  both  terms  of  the  fraction  f  by 

2>  i  x  \  =  TV  and>  if  \  =  T\>  T7  must  equal  I-  Hence, 
dividing  both  terms  of  a  fraction  by  the  same  number  does 
not  alter  its  value. 


ARITHMETIC. 


87.  A  fraction  is  reduced  to  its  lowest  terms  when  its 
numerator  and  denominator  cannot  both  be  divided  by  the 
same  number  without  a  remainder;  for  example,  f,  •§,  \\,  T87. 


88. 


EXAMPLES  FOR  PRACTICE 

Reduce  the  following : 
(a)     A  to  128ths. 


(<*) 


T^  to  its  lowest  terms. 
T§^  to  its  lowest  terms, 
f  to  49ths. 
U  to  lO.OOOths. 


Ans. 


89.  To  reduce  a  \vliole  number  or  a  mixed  number 
to  an  improper  fraction  : 

EXAMPLE. — How  many  fourths  in  5  ? 

SOLUTION.— Since  there  are  4  fourths  in  1  (f  ~  1),  in  5  there  will  be 
5X4  fourths,  or  20  fourths ;  i.  e. ,  5  X  |  =  --£•     Ans. 
EXAMPLE. — Reduce  8|  to  an  improper  fraction. 
SOLUTION.—    8  X  f-  =  -\2--     -£  +  f  =  -"/-•     Ans- 

90.  Rule. — Multiply  the  whole  number  by  the  denomina- 
tor of  the  fraction,  add  the  numerator  to  the  product  and 
place  the  denominator  under  the  result.      If  it  is  desired  to 
reduce  a  ivhole  number  to  a  fraction,  multiply  the  whole  number 
by  the  denominator  of  the  given  fraction,  and  write  the  result 
over  the  denominator. 


91. 


EXAMPLES  FOR  PRACTICE. 

Reduce  to  improper  fractions : 


51. 


(') 


37f. 
50*. 
Reduce  7  to  a  fraction  whose  denominator  is  16. 


Ans.  " 


92.  To  reduce  an  Improper  fraction  to  a  whole  or 
a  mixed  number: 

EXAMPLE. — Reduce  ^  to  a  mixed  number. 

SOLUTION. —  4  is  contained  in  21,  5  times  and  1  remaining  (see  Art. 
75) ;  as  this  is  also  divided  by  4,  its  value  is  J.  Therefore,  5  +  i,  or  5$, 
is  the  number.  Ans. 


26  ARITHMETIC.  §  1 

93.  Rule. — Divide  the  numerator  by  the  denominator, 
the  quotient  will  be  the  whole  number;   the  remainder,  if 
there  be  any,  ivill  be  the  numerator  of  the  fractional  part  of 
which  the  denominator  is  the  same  as  the  denominator  of  the 
improper  fraction. 

EXAMPLES  FOR  PRACTICE. 

94.  Reduce  to  whole  or  mixed  numbers: 


(«)      ***• 

(b)       if*. 


(d) 
(') 
(/)  W- 


(a)  241. 

(b)  61f. 

.        ,   (0   '  U«f 
ns''i   (ft)     49f. 
W      4. 

L  (/)    5. 


95.  A  common  denominator  of  two  or  more  fractions 
is  a  number  which  will  contain  (i.  e.,  which  may  be  divided 
by)  the  denominator  of  each  of  the  given  fractions  without 
a  remainder.     The   least   common   denominator  is  the 
least  number  that  will  contain  each   denominator  of   the 
given  fractions  without  a  remainder. 

96.  To  find  the  least  common  denominator: 

EXAMPLE. — Find  the  least  common  denominator  of  \,  \,  £,  and  TJff. 
SOLUTION. — We  first  place  the  denominators  in  a  row,  separated  by 
commas.  2  )  4,     3,     9,  16 

2)2,     3.     9,     8 

3  )  1,     3,     9.     4 

3  )  1,     1,     3,     4 

4  )  1.     1,     1,  ~4 

1,     1,     1,     1 
2X2X3X3X4  =  144,  the  least  common  denominator.     Ans. 

EXPLANATION. — Divide  each  of  them  by  some  prime  num- 
ber which  will  divide  at  least  two  of  them  without  a 
remainder  (if  possible),  bringing-  down  those  denominators 
to  the  row  below  which  will  not  contain  the  divisor  without 
a  remainder.  Dividing  each  of  the  numbers  by  2,  the  sec- 
ond row  becomes  2,  3,  9,  8,  since  2  will  not  divide  3  and  9 
without  a  remainder.  Dividing  again  by  2,  the  result  is  1, 
3,  9,  4.  Dividing  the  third  row  by  3,  the  result,  is  1,  1, 


§  1  ARITHMETIC.  27 

3,  4.  So  continue  until  the  last  row  contains  only  1's. 
The  product  of  all  the  divisors,  or  2  X  2  X  3  X  o  X  4  =  144,  is 
the  least  common  denominator. 

97.      EXAMPLE. — Find  the  least  common  denominator  of  |,  Tr^,  ^ 
SOLUTION.—  3  )  9,  12,  18 

3  )  3,     4,     6 

2  yi, 4^     2 


1 


1,     1^     1 

3X3X2X2  =  36.     Ans. 

98.  To  reduce  two  or  more  fractions  to  fractions 
having  a  common  denominator  : 

EXAMPLE.  —  Reduce  |,  £,  and  |  to  fractions  having  a  common  denom- 
inator. 

SOLUTION.  —  The  common  denominator  is  a  number  which  will  con- 
tain 3,  4,  and  2.  The  least  common  denominator  is  12,  because  it  is 
the  smallest  number  which  can  be  divided  by  3,  4,  and  2  without  a 
remainder.  2  _  R  3  _  9  t  __  „ 

3^   —    T?'       i    —    f?>       ?   —   T?- 

Reducing  f  (see  Art.  84),  3  is  contained  in  12,  4  times.  By  multi- 
plying both  numerator  and  denominator  of  f  by  4,  we  find 

^          =  :-j.     In  the  same  way  we  find  £  =  T9?  and  ^  =  TS5. 

99.  Rule.  —  Divide    tlic    common    denominator    by    the 
denominator  of  the  given  fraction,  and  multiply  both  terms 
of  the  fraction  by  the  quotient. 


EXAMPLES   FOU   PRACTICE. 

1OO.      Reduce  to  fractions  having  a  common  denominator: 

(/1\          867 
(a)        T'   5'   S- 


(C)      *•*•**'  Ans 

<<*)    f.  f-H- 
to     A.  A.  A- 


to 


w 


(/)    A,  H-  ii-  L   (/)    i*-  i^  It- 

ADDITION    OF    FRACTIONS. 

1O1.  Fractions  cannot  be  added  unless  they  have  a  com- 
mon denominator.  We  cannot  add  f  to  |  as  they  now  stand, 
since  the  denominators  represent  parts  of  different  sizes. 
Fourths  cannot  be  added  to  eighths. 


28  ARITHMETIC.  §  1 

Suppose  we  divide  an  apple  into  4  equal  parts,  and  then 
divide  2  of  these  parts  into  2  equal  parts.  It  is  evident 
that  we  shall  have  2  one-fourths  and  4  one-eighths.  Now, 
if  we  add  these  parts,  the  result  is  2  +  4  =  6  something. 
But  what  is  this  something  ?  It  is  not  fourths,  for  6 
fourths  are  1^,  and  we  had  only  1  apple  to  begin  with; 
neither  is  it  eighths,  for  6  eighths  are  f  ,  which  is  less  than 
1  apple.  By  reducing  the  quarters  to  eighths,  we  have 
|  =  |,  and  adding  the  other  4  eighths,  4  +  4  =  8  eighths. 
This  result  is  correct,  since  f  =  1.  Or  we  can,  in  this  case, 
reduce  the  eighths  to  quarters.  Thus,  |-  =  \  ;  whence, 
adding,  2  +  2  =  4  quarters,  a  correct  result,  since  \  =  1. 

Before  adding,  fractions  should  be  reduced  to  a  common 
denominator,  preferably  the  least  common  denominator. 

1O3.      EXAMPLE.  —  Find  the  sum  of  ^,  £,  and  f. 
SOLUTION.  —  The  least  common  denominator,  or  the  least  number 
which  will  contain  all  the  denominators,  is  8. 

l    —   4        s   —    6        anH      &   —    & 

T!  —  ~5>      f  —   S>      ana      •$  —   -5- 

EXPLANATION.  —  As  the  denominator  tells  or  indicates  the 
names  of  the  parts,  the  numerators  only  are  added,  to  obtain 
the  total  number  of  parts  indicated  by  the  denominator. 
Thus,  4  one-eighths  plus  6  one-eighths  plus  5  one-eighths  = 

««        » 


An, 


1O3.      EXAMPLE.—  What  is  the  sum  of  12f,  14f,  and  7Ty 
SOLUTION.  —  The  least  common  denominator  in  this  case  is  16. 
12f  =  123-f 

14f  = 


__ 

sum    83  +  f*  =  33  +  111  =  3411, 

The  sum  of  the  fractions  =  f  \  or  \\\,  which  added  to  the  sum  of  the 
whole  numbers  = 


EXAMPLE.  —  What  is  the  sum  of  17,  13T8^,  -fa,  and  3J? 

SOLUTION.  —  The  least  common   denominator  is    32.     13T\  =  1358¥, 

=  3£.  17 


sum    33ff.     Ans. 


§  1  ARITHMETIC.  20 

1O4.  Rule. — I.  Reduce  the  given  fractions  to  frac- 
tions having  the  least  common  denominator,  and  write  the 
sum  of  the  numerators  over  the  common  denominator. 

II.  When  there  are  mixed  numbers  and  whole  numbers, 
add  the  fractions  first,  and  if  tlicir  sum  is  an  improper 
fraction,  reduce  it  to  a  mixed  number  and  add  the  whole 
number  witJi  the  other  whole  numbers. 


EXAMPLES  FOR  PRACTICE. 

1O5.      Find  the  sum  of: 


(f)     4.  f >  A- 

V*)       £>   T5>   if-  A 

Ans. 

(/)    ff.tt.il 
(/•)    A.A-*i 


SUBTRACTION    OF    FRACTIONS. 

106.  Fractions  cannot  be  subtracted  without  first   re- 
ducing   them    to    a    common    denominator.       This   can    be 
shown  in  the  same  manner  as  in  the  case  of  addition  of 
fractions. 

EXAMPLE. — Subtract  f  from  |f. 

SOLUTION. — The  common  denominator  is  16. 

I  =  A-     it-A=    ^jp    =A-    Ans. 

107.  EXAMPLE. — From  7  take  f. 

SOLUTION. —    1  =  f ;  therefore,  since  7  =  6  +  1,  7  =  6  +  f  =  6£,  or 
6f-i  _=  6f.     Ans. 

108.  EXAMPLE. — What  is  the  difference  between  17Tflff  and  9-J-f  ? 
SOLUTION. — The  common  denominator  of  the  fractions  is  32.     17-j*s 

=  17«.  .          ,    .... 

minuend    17if 

subtrahend      9^| 
difference      8^. 


30  ARITHMETIC.  §  1 

1O9.      EXAMPLE.— From  9£  take  4^. 

SOLUTION. — The  common   denominator  of  the  fractions  is  16.     9£ 

-  9iV  minuend    9T4S  or  8f£  - 

subtrahend  .4T7ff       4T7ff 
difference    4|f       4^|.     Ans. 

EXPLANATION. — As  the  fraction  in  the  subtrahend  is 
greater  than  the  fraction  in  the  minuend,  it  cannot  be  sub- 
tracted; therefore,  borrow  1,  or  if,  from  the  9  in  the 
minuend  and  add  it  to  the  T4g-;  yV  +  yf  =  yf-  TV  from 
|f  =  if.  Since  1  was  borrowed  from  9,  8  remains ;  4  from 


8  =  4;  4  +  H  =  4|f- 

HO.      EXAMPLE. — From  9  take  8Tsff. 

SOLUTION. —  minuend    9     or 

subtrahend    8A 


difference      \\        \\.     Ans. 

EXPLANATION. — As  there  is  no  fraction  in  the  minuend 
from  which  to  take  the  fraction  in  the  subtrahend,  borrow 
1,  or  if,  from  9.  f\  from  if  =  ^f .  Since  1  was  borrowed 
from  9,  only  8  is  left.  8  from  8  =  0. 

111.  Rule. — I.  Reduce  the  fractions  to  fractions 
having  a  common  denominator.  Subtract  one  numerator 
from  the  other  and  place  the  remainder  over  the  common 
denominator. 

II.  When  there  are  mixed  numbers,  subtract  the  fractions 
and  whole  numbers  separately,  and  place  the  remainders  side 
by  side. 

III.  When   the  fraction   in   the  subtrahend  is   greater 
than  the  fraction  in  the  minuend,  borrow  1  from  the  whole 
number   in    the    minuend  and  add    it   to   the  fraction    in 
the  minuend,  from  which  subtract  the  fraction  in  the  sub- 
trahend. 

IV.  When  the  minuend  is  a  whole  number,  borrow  1; 
reduce  it  to  a  fraction  whose  denominator  is  the  same  as  the 
denominator  of  the  fraction  in  the  subtrahend,  and  place  it 
over  that  fraction  for  subtraction. 


ARITHMETIC.  31 


EXAMPLES  FOR  PRACTICE. 
112.      Subtract: 

(a)  ^  from  f|. 

(b)  T7T  from  ||. 

(c)  A,  from  T5g. 
(<i)     if  from  ig. 

(*)      }£  from  ff.  AnS<   1 

(_/")     13^  from  30|. 
(£•)     12i  from  27. 


(//)      5i  from  30. 


(a) 
(t) 

(JO 


I  (//)      2-H. 


MUI/riPLICATIOX    OF    FRACTIONS. 

113.  In  multiplication  of  fractions  it  is  not  necessary  to 
reduce  the  fractions  to  fractions  having  a  common  denominator. 

114.  Multiplying  the  numerator  or  dividing  the  denom- 
inator multiplies  the  fraction. 

EXAMPLE.— Multiply  £  by  4. 

3  V  4 
SOLUTION.—  |x4  =  ^         =  1*  =  3.    Ans. 

4 

Or,     IX4  =  -  f  =  3.     Ans. 


The  word  "of,"  when  placed  between  two  fractions,  or 
between  a  fraction  and  a  whole  number,  means  the  same  as 
X ,  or  times.  Thus, 

|  of  4  =  |X4  =  3. 

8  Ol  T6"   "    8  X  y^    "   TT¥- 
EXAMPLE. — Multiply  f  by  2. 

3x2 
SOLUTION. —  2X1  =  77         =  |  =  f  •     A"S. 

Or,     2X1  =  |      o  =  f  •     Ans. 

O    -T-    « 

115.      EXAMPLE. — What  is  the  product  of  T4ff  and  |  ? 

SOLUTION.—       ^X|  =  «2a==Wk  =  A-     Ans- 
lox  o 

Or,  by  cancelation,        y^^  =  3-^g  =  ^.     Ans. 


4 

116.     EXAMPLE.— What  is  f  of  f  of  £|  ? 
SOLUTION.—  *  X  '!  X  ff  =  — i-,  =  A.     Ans. 


J  ARITHMETIC. 

117,      EXAMPLE. — What  is  the  product  of  9f  and  5|  ? 
SOLUTION.—  9f  =  *£;  5|  =  -4/. 

on  \/  Ai 

Ans. 


118.      EXAMPLE.—  Multiply  15|  by  3. 
SOLUTION.  —  15£  15| 

3     or         3 


47f  45  +  -V-  =  45  +  2f  =  47f.     Ans. 

119.  Rule.  —  I.  Divide  t  lie  product  of  the  numerators 
by  the  product  of  the  denominators.  All  factors  common  to 
the  numerators  and  denominators  should  first  be  cast  out  by 
cancelation. 

II.  To  multiply  one  mixed  number  by  another,  reduce  them 
both  to  improper  fractions. 

III.  To  multiply  a  mixed  number  by  a  whole  number  ',  first 
multiply  the  fractional  part  by  the  multiplier,  and  if  the  prod- 
uct is  an  improper  fraction,  reduce  it  to  a  mixed  number 
and  add  the  whole-number  part  to  the  product  of  the  mul- 
tiplier and  the  whole  number. 


EXAMPLES  FOB  PRACTICE. 

1 2O.      Find  the  product  of : 
(«)     7  X  TV 
(*)      14  X  A- 


<<0 

(/) 


-X7  AnS'  J 

Tff  X   <• 


iff  X  32. 

ifX!4. 


(/)     125. 


DIVISION    OF   FRACTIONS. 

121.  /«  division  of  fractions  it  is  not  necessary  to  reduce 
the  fractions  to  fractions  having  a  common  denominator. 

122.  Dividing  the  numerator  or  multiplying  the  denom- 
inator divides  the  fraction. 

EXAMPLE. — Divide  f  by  3. 

SOLUTION. — When  dividing  the  numerator,  we  have 

H'3  =      =    .     Ans. 


1  ARITHMETIC.  33 

When  multiplying  the  denominator,  we  have 

A     .     '\    —  -      fi      __    1  A  n  c 

-  8  X  8       ™  ~  *'     AnS> 
EXAMPLE.  —  Divide  -£f  by  2. 
SOLUTION.—  ft  -  2  =  ^  ^  g  =  &.     Ans. 

EXAMPLE.  —  Divide  i|  by  7. 

1  4.  -^-  T 
SOLUTION.  —        if  -r-  7  =    ^  '       =  ^  =  -^     Ans. 


123.  To  invert  a  fraction  is  to  /wr«  //  upside  doivn  ; 
that  is,  make  the  numerator  and  denominator  change  places. 

Invert  f  and  it  becomes  |. 

124.  EXAMPLE.—  Divide  T9ff  by  T\. 

SOLUTION.  —  1.  The  fraction  T3g-  is  contained  in  ^,  3  times,  for  the 
denominators  are  the  same,  and  one  numerator  is  contained  in  the 
other  3  times.  2.  If  we  now  invert  the  divisor,  Tsff,  and  multiply,  the 
solution  is 

3 
9       16        ?  X  W        o       A 

T6xT  =  ;F"x^=    '   Ans' 

This  brings  the  same  quotient  as  in  the  first  case. 

125.  EXAMPLE.  —  Divide  f  by  \. 

SOLUTION.  —  We  cannot  divide  f  by  J,  as  in  the  first  case  above,  for 
the  denominators  are  not  the  same  ;  therefore,  we  must  solve  as  in  the 
second  case. 

3  V  4        3 


Ans- 


2 

126.      EXAMPLE.—  Divide  5  by  |f. 
SOLUTION.  —         inverted  becomes 


127.  EXAMPLE.  —  How  many  times  is  3|  contained  in  7T7j-? 

SOLUTION.—  3f  =  ^;  7rV  =  ¥/• 

•^  inverted  equals  T4?. 

119       4        119  X^       119 
-  X  —  =  -  ~  =  --  =  1|&.     Ans. 
16        15       }$  X  15        60 

4 

128.  Rule.  —  Invert  the  divisor  and  proceed  as  in  mul- 
tiplication. 


34  ARITHMETIC.  §  1 

129.  We  have  learned  that  a  line  placed  between  two 
numbers  indicates  that  the  number  above  the  line  is  to  be 
divided  by  the  number  below  it.  Thus,  -L8-  shows  that  18 
is  to  be  divided  by  3.  This  is  also  true  if  a  fraction  or  a 
fractional  expression  be  placed  above  or  below  a  line. 

9  3x7 

-  means  that  9  is  to  be  divided  by  f ;  means  that 


3  X  7  is  to  be  divided  by  the  value  of 
•   is  the  same  as     -j-   . 


16 

8  +  4 
16   ' 


It  will  be  noticed  that  there  is  a  heavy  line  between  the  9 
and  the  £ .  This  is  necessary,  since  otherwise  there  would  be 
nothing  to  show  as  to  whether  9  was  to  be  divided  by  f ,  or 
-§-  was  to  be  divided  by  8.  Whenever  a  heavy  line  is  used, 
as  shown  here,  it  indicates  that  all  above  the  line  is  to  be 
divided  by  all  bcloiv  it. 


13O. 


EXAMPLES   FOK  PRACTICE. 

Divide 

(a) 

15  by  6f  . 

'(«) 

(b) 

30  by  f  . 

(*) 

W 

172  by  *. 

W 

(d) 
(*) 

los  b    142                        Ans.  - 

W 

(/) 

Aj4/  by  17$. 

(/: 

te) 

rl  b7  TTT- 

(^ 

(*) 

\8/  by  72f 

.  w 

40. 
215. 

ttf 

HI- 


131.  Whenever  an  expression  like  one  of  the  three 
following'  ones  is  obtained,  it  may  always  be  simplified  by 
transposing  the  denominator  from  above  to  below  the  line, 
or  from  below  to  above,  as  the  case  may  be,  taking  care, 
however,  to  indicate  that  the  denominator  when  so  trans- 
ferred is  a  multiplier. 

*  3 

=  ^5    for,    regarding    the    fraction 


9X4 
above  the  heavy  line  as  the  numerator  of  a  fraction  whose 

1X4  3 

denominator  is  9,  i         =  - — -,  as  before, 
y  x  ^       j  x  4 


§  1  ARITHMETIC.  35 

9        9x4 

2.  -r  =  —  ^—  =  12.     The  proof  is  the  same  as  in  the  first 
f  o 

case. 

-        5X4 

3.  -|  =  -  —  -  =  |f     For,  regarding-  f  as  the  numerator 

^£  O  X  J 

—  X  9  5 

of  a  fraction  whose  denominator  is  f  ,  -|     '    =  -  —  -  ;    and 

•j  X  J         o  X  J 

5     X4        5X4  4 

-  —  -        =  -  —  -  =  |4,  as  above. 
3X9  3X9        2" 

~~T~ 

This  principle  may  be  used  to  great  advantage  in  cases 


V1  _    .      . 

like  -  —  —  -  —  .  .  '  '    .  —  .     Reducing    the    mixed    numbers   to 
40  X  4£  X  51 


t      *.•          4.1,  •        i 

fractions,  the  expression    becomes  -  —  —  —    -^-.  -  .      Now 


transferring  the  denominators  of  the  fractions  and  canceling, 

3 

;p    3     p        3 

1x310x27x72x2x6  _    lx;S;px  £7x^x2x0 
40X9X31X4X12  ^px))X^x4x^ 

i  2 

=  ^  =  134  * 

2 

Greater  exactness  in  results  can  usually  be  obtained  by 
using  this  principle  than  by  reducing  the  fractions  to  deci- 
mals. The  principle,  however,  should  not  be  employed  if 
a  sign  of  addition  or  subtraction  occurs  either  above  or 
below  the  dividing  line. 


DECIMALS. 

132.  Decimals  are  tenth  fractions;  that  is,  the  parts  of 
a  unit  are  expressed  on   the   scale  of  ten,  as  tenths,  hnn- 
drcdths,  thousandths,  etc. 

133.  The  denominator  which  is  always  ten  or  a  multiple 
of  ten,  as  10,  100,  1,000,  etc.,  is  not  expressed,  as  it  would 


36  ARITHMETIC.  §  1 

be  in  common  fractions,  by  writing  it  under  the  numerator 
with  a  line  between  them,  as  T3¥,  Tf ¥,  T7\7,  but  is  expressed 
by  placing  a  period  ( .),  which  is  called  a  decimal  point,  to 
the  left  of  the  figures  of  the  numerator,  so  as  to  indicate 
that  the  number  on  the  right  is  the  numerator  of  a  fraction 
whose  denominator  is  10,  100,  1,000,  etc. 

134.  The  reading  of  a  decimal  number  depends  upon 
the  number  of  decimal  places  in  it,  or  the  number  of  figures 
to  the  right  of  the  decimal  point. 

One  decimal  place  expresses  tenths. 

Two  decimal  places  express  hundredths. 

Three  decimal  places  express  thousandths. 

Four  decimal  places  express  ten-tJiousandths. 

Five  decimal  places  express  Jiundred-thousandths. 

Six  decimal  places  express  millionths. 

=  3  tenths. 
=  3  hundredths. 
=  3  thousandths. 
=  3  ten-thousandths. 
=  3  hundred-thousandths. 
j.    =  3  millionths. 

We  see  in  the  above  that  the  number  of  decimal  places  in 
a  decimal  equals  the  number  of  ciphers  to  the  right  of  the 
figure  1  in  the  denominator  of  its  equivalent  fraction.  This 
fact  kept  in  mind  will  be  of  much  assistance  in  reading  and 
writing  decimals. 

Whatever  may  be  written  to  the  left  of  a  decimal 
point  is  a  whole  number.  The  decimal  point  merely  sep- 
arates the  fraction  on  the  right  from  the  whole  number  on 
the  left. 

When  a  whole  number  and  decimal  are  written  together, 
the  expression  is  a  mixed  number.  Thus,  8. 12  and  17.25  are 
mixed  numbers. 


§  1  ARITHMETIC.  37 

The  relation  of  decimals  and  whole  numbers  to  each  other 
is  clearly  shown  by  the  following  table : 

co' 

'O 

I    L  I    i 

*§  *!        ~     .11  ,-i 

<4_i^  VM       3  .G  •       W       2       O  'til     rG 

«    !      .     «    !    T3     «  a         £    I     g    V     «    J     £ 

<u   M-I     g     o>    M-I     S     0  13      •     o    S          o    "£   ri:     o 

VH  0s-1  So        «-!  •        ^        W        *-<        $      ,G        <-<        O        3       *-< 

*°     r/^-S^      ^      3    T3       •      w      G    X    ^(      3    -3    ^    .2      £    -3 

,-iCO> — ir-iCOi->_itO-t-i-'-i4_i^,-J          i          ^^_          ,          _ 

gG^SCogc-gOrHGoGS^GS 
^cup;2(3JTG^!u5rS:on;TGo^co^ 
^d^-)  G^q^-i-M^^  Grd-i-i^+-i-i-i^  G-urG 

987654321     .      23456789 

The  figures  to  the  left  of  the  decimal  point  represent 
whole  numbers;  those  to  the  right  are  decimals. 

In  both  the  decimals  and  whole  numbers,  the  units  place 
is  made  the  starting  point  of  notation  and  numeration. 
Both  whole  numbers  and  decimals  decrease  on  the  scale  of 
ten  to  the  right,  and  both  increase  on  the  scale  of  ten  to  the 
left.  The  first  figure  to  the  left  of  units  is  tens,  and  the 
first  figure  to  the  right  of  units  is  tenths.  The  second  figure 
to  the  left  of  units  is  hundreds,  and  the  second  figure  to  the 
right  is  hundredths.  The  third  figure  to  the  left  is  thousands, 
and  the  third  to  the  right  is  thousandths,  and  so  on ;  the 
w/io/e  numbers  on  the  left  and  the  decimals  on  the  right. 
The  figures  equally  distant  from  units  place  correspond  in 
name,  the  decimals  having  the  ending  ths,  to  distinguish 
them  from  whole  numbers.  The  following  is  the  numeration 
of  the  number  in  the  above  table :  nine  hundred  eighty-seven 
million,  six  hundred  fifty-four  thousand,  three  hundred 
twenty-one  and  twenty-three  million,  four  hundred  fifty-six 
thousand,  seven  hundred  eighty-nine  hundred-millionths. 

The  decimals  increase  to  the  left,  on  the  scale  of  ten,  the 
same  as  whole  numbers;  for,  if  you  begin  at  the  4  in 
thousandths  place  in  the  above  table,  the  next  figure  to  the 
left  is  hundredths,  which  is  ten  times  as  great,  and  the  next 
tenths,  or  ten  times  the  hundredths,  and  so  on  through  both 
decimals  and  whole  numbers. 


38  ARITHMETIC.  §  1 

1  35.     Annexing,  or  taking  azvay,  a  cipher  at  the  right  of 
a  decimal,  does  not  affect  its  value. 

.5  is  T*T;  .50  is  ^,  but  fV  =  T5oV,  therefore,  .5  =.50. 

136.  Inserting  a  cipher  between  a  decimal  and  the  decimal 
point,  divides  the  decimal  by  10. 

K-.       5.       5       -in    —        5        —     AK 

•  °    —   TIT  '    TO  ~  ±       —   TOTT   --  uo- 

137.  Taking  away  a  cipher  from  the  left  of  a  decimal, 
multiplies  the  decimal  by  10. 


_  5.         5      v  1  0    —      5      — 

—   TOTT'    To¥*11        -   TIT   -- 


138.  In  some  cases  it  is  convenient  to  express  a  mixed 
decimal  fraction  in  the  form  of  a  common  (improper)  frac- 
tion. To  do  so  it  is  only  necessary  to  write  the  entire  num- 
ber, omitting  the  decimal  point,  as  the  numerator  of  the 
fraction,  and  the  denominator  of  the  decimal  part  as  the 
denominator  of  the  fraction.  Thus,  127.483  =  J-fUt1;  for, 
127.483  =  1 


ADDITION    OF    DECIMALS. 

139.  Addition  of  decimals  is  similar  in  all  respects  to 
addition  of  whole  numbers — units  are  placed  under  units, 
tens  under  tens,  etc. ;  this,  of  course,  brings  the  decimal 
points  in  line,  directly  under  one  another.  Hence,  in  pla- 
cing the  numbers  to  be  added,  it  is  only  necessary  to  take 
care  that  the  decimal  points  are  in  line.  In  adding  whole 
numbers,  the  right-hand  figures  are  always  in  line;  but 
in  adding  decimals,  the  right-hand  figures  will  not  be  in 
line  unless  each  decimal  contains  the  same  number  of 
figures. 

wliole  numbers  decimals  mixed  numbers 

342  .342  342.032 

4234  .4234  4234.5 

26  .26  26.6782 

3^  .03  3.06 

sum    4605    Ans.  sum    1.0554  Ans.      sum    4606.2702    Ans. 


§1 


ARITHMETIC. 


39 


140.  EXAMPLE.— What  is  the  sum  of  242,  .36,  118.725,   LOOS,  G, 
and  100.1? 

SOLUTION. —  242. 

.30 

1  1  8.7  2  5 
1.005 
6. 

1  00.1 
sum     4  6  8. 1  9  0     Ans. 

141.  Jlule. — Place  the  numbers  to  be  added  so  that  t/ie 
decimal  points  will  be  directly  under  each  other.     Add  as 
in  whole  numbers,  and  place  the  decimal  point  in  the  sum. 
directly  under  the  decimal  points  above. 


(a) 
(£) 
(C) 


(e) 
(/) 
(g) 
(h) 


EXAMPL/ES  FOR  PRACTICE. 

14/2.      Find  the  sum  of: 

.2143,  .105,  2.3042,  and  1.1417. 
783.5,  21.478,  .2101,  and  .7816. 
21.781,  138.72,  41.8738,  .72,  and  1.413. 
.3724,  104.15,  21.417,  and  100.042. 
200.172,  14.105,  12.1465,  .705,  and  7.2. 
1,427.16,  .244,  .32,  .032,  and  10.0041. 

2.473.1,  41.65,  .7243,  104.067,  and  21.073. 

4.107.2,  .00375,  21.716,  410.072,  and  .0345. 


Ans. 


•  (a) 
(*) 

('0 

kr) 

3.7652. 
805.9647. 
204.507S. 
225.9814. 
234.3285. 
1,437.7601. 
2,640.6143. 
4,539.02625. 

SUBTRACTION    OF    DFCIMAI^S. 

143.  As  in  subtraction   of  whole  numbers,    units   are 
placed   under   units,    tens   under   tens,    etc.,    bringing    the 
decimal  points  under  each  other,  as  in  addition  of  decimals. 

EXAMPLE.— Subtract  .132  from  .3063. 
SOLUTION. —  minuend    .3063 

subtrahend    .1  32 
difference    .1743    Ans. 

144.  EXAMPLE.— What  is  the  difference  between  7.895  and  .725  ? 
SOLUTION. —  minuend    7.8  9  5 

subtrahend      .725 
difference    7.1  7  0  or  7.1  7.     Ans. 


40  ARITHMETIC.  §  1 

145.  EXAMPLE.— Subtract  .625  from  11. 
SOLUTION. —  minuend    1  1.0  0  0 

subtrahend         .625 
difference     1  0.3  7  5    Ans. 

146.  Rule.—  Place  the  subtrahend  under  the  minuend, 
so  that  the  decimal  points  will  be  directly  under  each  other. 
Subtract  as  in  whole  numbers,  and  place  the  decimal  point  in 
the  remainder  directly  under  the  decimal  points  above. 

When  the  figures  in  the  decimal  part  of  the  subtraJicnd 
extend  beyond  those  in  the  minuend,  place  ciphers  in  the  min- 
uend above  them  and  subtract  as  before. 


EXAMPLES  FOR  PRACTICE. 
147.      From: 


(a)  407.385  take  235.0004. 

(&)  22. 718  take  1.7042. 

(c)  1,368. 17  take  13.6817. 

(d)  70.00017  take  7.000017. 

(e)  630.630  take  .6304. 
(/)  421.73  take  217. 162. 
(g)  1.000014  take  .00001. 
(h)  .783652  take  .542314. 


Ans. 


(a)  172.3846. 

(b)  21.0138. 

(c)  1,354.4883. 

(d)  63.000153. 

(e)  629.9996. 
(/)  204.568. 
(g)  1.000004. 

6  .241338. 


MULTIPLICATION    OF    DECIMALS. 

148.  In  multiplication  of  decimals  we  do  not  place  the 
decimal  points  directly  under  each  other  as  in  addition  and 
subtraction.  We  pay  no  attention  for  the  time  being  to  the 
decimal  points.  Place  the  multiplier  under  the  multiplicand, 
so  that  the  right-hand  figure  of  the  one  is  under  the  right- 
hand  figure  of  the  other,  and  proceed  exactly  as  in  multipli- 
cation of  whole  numbers.  After  multiplying,  count  the 
number  of  decimal  places  in  both  multiplicand  and  multiplier, 
and  point  off  the  same  number  in  the  product. 

EXAMPLE.— Multiply  .825  by  13. 
SOLUTION. —       multiplicand         .825 
multiplier  1  3 

2475 

825 


product    1  0.7  2  5    Ans. 


§  1  ARITHMETIC.  41 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plicand and  none  in  the  multiplier;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

149.  EXAMPLE.— What  is  the  product  of  426  and  the  decimal  .005  ? 
SOLUTION. —        multiplicand       426 

multiplier       .00  5 

product     2.1  30  or  2. 13.     Ans. 

In  this  example  there  are  3  decimal  places  in  the  mul- 
tiplier and  none  in  the  multiplicand;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

150.  It  is  not  necessary  to  multiply  by  the  ciphers  on 
the  left  of  a  decimal;  they  merely  determine  the  number 
of  decimal  places.      Ciphers  to  the  right  of  a  decimal  should 
be  omitted,  as  they  only  make  more  figures  to  deal  with, 
and  do  not  change  the  value. 

151.  EXAMPLE.— Multiply  1.205  by  1.15. 
SOLUTION. —     multiplicand          1.2  0  5 

multiplier  1.1  5 

6025 
1205 
1205 


product     1.38575     Ans. 

In  this  example  there  are  3  decimal  places  in  the  mul- 
tiplicand and  2  in  the  multiplier ;  therefore,  3  +  2,  or  5, 
decimal  places  must  be  pointed  off  in  the  product. 

152.  EXAMPLE.— Multiply  .232  by  .001. 
SOLUTION. —     multiplicand  .232 

multiplier  .001 

product    .000232    Ans. 

In  this  example  we  multiply  the  multiplicand  by  the  digit 
in  the  multiplier,  which  gives  232  for  the  product;  but  since 
there  are  3  decimal  places  each  in  the  multiplier  and  multi- 
plicand, we  must  prefix  3  ciphers  to  the  232  to  make  3  -{-  3, 
or  6,  decimal  places  in  the  product. 

1 53.  Rule. — Place  the  multiplier  under  the  multiplicand, 
disregarding  the  position  of  the  decimal  points.     Multiply 


42  ARITHMETIC.  §  1 

as  in  whole  numbers,  and  in  the  product  point  off  as  many 
decimal  places  as  there  are  decimal  places  in  both  multiplier 
and  multiplicand,  prefixing  ciphers  if  necessary. 


EXAMPLES  FOR  PRACTICE. 

154.      Find  the  product  of: 


(a)  .000492X4.1418. 

(b)  4,003.2X1.2. 
(<;)  78.6531x1-03. 
(rf)  .3685  X. 042. 
(e)  178,352  X  .01. 
(/)  .  00045  X- 0045. 
(g)  .714  X  .00002. 
(h)  .  00004  X- 008. 


(a)  .0020377656. 

(6)  4,803.84. 

(c)  81.012693. 

(d)  .015477. 

(e)  1,783.52. 
(/)  .000002025. 
(g)  .00001428. 
(h)  .00000032. 


DIVISION    OF    DECIMALS. 

155.  In  division  of  decimals  we  pay  no  attention  to  the 
decimarpoint  until  after  the  division  has  been  performed. 
The  number  of  decimal  places  in  the  dividend  must  equal  (or 
be  made  to  equal  by  annexing  ciphers]  the  number  of  decimal 
places  in  the  divisor.  Divide  exactly  as  in  whole  numbers. 
Subtract  the  number  of  decimal  places  in  the  divisor  from  the 
number  of  decimal  places  in  the  dividend,  and  point  off  as 
many  decimal  places  in  the  quotient  as  there  are  units  in  the 
remainder  thus  found. 

EXAMPLE.— Divide  .625  by  25. 

divisor  dividend  quotient 

SOLUTION.—  2  5  )  .6  2  5  ( .0  2  5    Ans. 

50 
125 
125 
remainder       0 

In  this  example  there  are  no  decimal  places  in  the  divisor, 
and  three  decimal  places  in  the  dividend ;  therefore,  there 
are  3  minus  0,  or  3,  decimal  places  in  the  quotient.  One 
cipher  has  to  be  prefixed  to  the  25  to  make  the  three  decimal 
places. 


ARITHMETIC.  43 


15G.      EXAMPLE.— Divide  6.035  by  .05. 

divisor   dividend   quotient 

SOLUTION.—  .05)6.035(120.7    Ans. 

5 

1  0 
1  0 


35 
35 

remainder     0 

In  this  example  we  divide  by  5,  as  if  the  cipher  were  not 
before  it.  There  is  one  more  decimal  place  in  the  dividend 
than  in  the  divisor ;  therefore,  one  decimal  place  is  pointed 
off  in  the  quotient. 

157.  EXAMPLE.— Divide  .125  by  .005. 

divisor  dividend  quotient 

SOLUTION.—  .005). 125(25    Ans. 

1  0 

25 

25 

remainder    0 

In  this  example  there  are  the  same  number  of  decimal 
places  in  the  dividend  as  in  the  divisor ;  therefore,  the  quo- 
tient has  no  decimal  places,  and  is  a  whole  number. 

158.  EXAMPLE.— Divide  326  by  .25. 

divisor  dividend     quotient 

SOLUTION.—  .2  5)32  6. 00(1304    Ans. 

25 

76 

75 


100 


remainder        0 


In  this  problem  two  ciphers  were  annexed  to  the  div- 
idend, to  make  the  number  of  decimal  places  equal  to 
the  number  in  the  divisor.  The  quotient  is  a  whole 
number. 


44  ARITHMETIC.  §  1 

159.  EXAMPLE.— Divide  .0025  by  1.25. 

divisor      dividend    quotient 
SOLUTION.—  1.2  5  )  .0  0  2  5  0  ( .0  0  2    Ans. 

250 
remainder        0 

EXPLANATION. — In  this  example  we  are  to  divide  .0025  by 
1.25.  Consider  the  dividend  as  a  whole  number,  i.  e.,  as  25 
(disregarding  the  two  ciphers  at  its  left,  for  the  present) ; 
also,  consider  the  divisor  as  a  whole  number,  i.  e. ,  as  125. 
It  is  clearly  evident  that  the  dividend,  25,  will  not  contain 
the  divisor,  125;  we  must,  therefore,  annex  one  cipher  to 
the  25,  thus  making  the  dividend  250.  125  is  contained 
twice  in  250,  so  we  place  the  figure  2  in  the  quotient.  In 
pointing  off  the  decimal  places  in  the  quotient,  it  must  be 
remembered  that  there  were  only  four  decimal  places  in  the 
dividend;  but  one  cipher  was  annexed,  thereby  making 
4  +  1,  or  5,  decimal  places.  Since  there  are  five  decimal 
places  in  the  dividend  and  two  decimal  places  in  the  divisor, 
we  must  point  off  5  —  2,  or  3,  decimal  places  in  the  quotient. 
In  order  to  point  off  three  decimal  places,  two  ciphers  must 
be  prefixed  to  the  figure  2,  thereby  making  .002  the  quo- 
tient. It  is  not  necessary  to  consider  the  ciphers  at  the  left 
of  a  decimal  when  dividing,  except  when  determining  the 
position  of  the  decimal  point  in  the  quotient. 

160.  Rule. — I.    Place  the  divisor  to  the  left  of  the  divi- 
dend, and  proceed  as  in  division  of  whole  numbers;  in  the 
quotient,  point  off  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceed  those  in  the  divisor,  pre- 
fixing ciphers  to  the  quotient,  if  necessary. 

II.  If  in  dividing  one  number  by  another  there  be  a 
remainder,  the  remainder  can  be  placed  over  the  divisor, 
as  a  fractional  part  of  the  quotient,  but  it  is  generally 
better  to  annex  ciphers  to  the  remainder,  and  continue 
dividing  until  there  are  3  or  4  decimal  places  in  the  quo- 
tient, and  then  if  there  still  be  a  remainder,  terminate  the 
quotient  by  the  plus  sign  (+),  which  shows  that  it  can  be 
carried  further. 


1  ARITHMETIC.  45 

161.      EXAMI-LK.—  What  is  the  quotient  of  199  divided  by  15  ? 
divisor  dividend  quotient 

SOLUTION.—  15)199(13  + T45    Ans. 

1  5 
49 

remainder    4 

Or,     15)199.000(13.266+     Ans. 
1  5 

49 
45_ 

40 

30 


1  00 
90_ 
100 
90 

remainder     1  0 
13TV  =  13.266  + 
&  =       -266  + 

!(>*£.  It  frequently  happens,  as  in  the  above  example, 
that  the  division  will  never  terminate.  In  such  cases,  decide 
to  how  many  decimal  places  the  division  is  to  be  carried, 
and  carry  the  work  one  place  further.  If  the  last  figure  of 
the  quotient  thus  obtained  is  5  or  a  greater  number,  increase 
the  preceding  figure  by  1,  and  write  after  it  the  minus  sign 
(— ),  thus  indicating  that  the  quotient  is  not  quite  as  large 
as  indicated ;  if  the  figure  thus  obtained  is  less  than  5,  write 
the  plus  sign  (-(-)  after  the  quotient,  thus  indicating  that 
the  number  is  slightly  greater  than  as  indicated.  In  the 
last  example,  had  it  been  desired  to  obtain  the  answer  cor- 
rect to  four  decimal  places,  the  work  would  have  been  car- 
ried to  five  places,  obtaining  13.2G66G,  and  the  answer  would 
have  been  given  as  13.2667  —  .  This  remark  applies  to  any 
other  calculation  involving  decimals,  when  it  is  desired  to 
omit  some  of  the  figures  in  the  decimal.  Thus,  if  it  was  de- 
sired to  retain  three  decimal  places  in  the  number  .2471253, 
it  would  be  expressed  as  .247  +  ;  if  it  was  desired  to  retain 
five  decimal  places,  it  would  be  expressed  as  .24713  —  . 
Both  the  +  and  —  signs  are  frequently  omitted ;  they  are 


46 


ARITHMETIC. 


seldom  used  outside  of  arithmetic,  except  in  exact  calcula- 
tions, when  it  is  desired  to  call  particular  attention  to  the 
fact  that  the  result  obtained  is  not  quite  exact. 


EXAMPLES  FOB  PRACTICE. 


163.      Divide: 

(a)      101.6688  by  2.36. 
(6)      187.12264  by  123.107. 

.08  by  .008. 

.0003  by  3.75. 

.0144  by  .024. 

.00375  by  1.25. 

.004  by  400. 

.4  by  .008. 


(d) 

w 


Ans. 


w 


(') 


43.08. 

1.52. 

10. 

.00008. 

.6. 

.003. 

.00001. 

50. 


REDUCTION    OF    DECIMALS. 


TO  REDUCE  A  FRACTION  TO  A  DECIMAL. 

164.      EXAMPLE.  —    f  equals  what  decimal  ? 


SOLUTION.— 


M|or.=,, 


Ans. 


EXAMPLE. — What  decimal  is  equivalent  to 
SOLUTION.—  8  )  7.0  0  0  (  .8  7  5 

64 


60 
56 

"To 

40 

0 


or      =  .875.     Ans. 


165.  Rule. — Annex  ciphers  to  the  numerator  and  divide 
by  the  denominator.  Point  off  as  many  decimal  places  in  the 
quotient  as  there  are  ciphers  annexed. 


166. 


EXAMPLES  FOR  PRACTICE. 

Reduce  the  following  common  fractions  to  decimals: 


(a) 
(*) 


(f) 

(K) 


if- 
I- 

H- 
U- 
A- 


1000' 


Ans.  • 


(«) 


.46875. 

.875. 

.65625. 

.796875. 

.16. 

.625. 

.05. 

.004. 


§  1  ARITHMETIC.  47 

167.  To  reduce  inches  to  decimal  parts  of  a  foot  : 

EXAMPLE. — What  decimal  part  of  a  foot  is  9  inches  ? 

SOLUTION. — Since  there  are  12  inches  in  one  foot,  1  inch  is  ^  of  a 
foot,  and  9  inches  is  9  X  TJ>  or  iV  °f  a  f°°t.  This  reduced  to  a  decimal 
by  the  above  rule  shows  what  decimal  part  of  a  foot  9  inches  is. 

1  2)  9.00  (.7  5  of  a  foot.     Ans. 

84 

GO 

<H) 

0 

168.  Rule. — I.      To  reduce  indies  to  a  decimal  part  of 
a  foot,  divide  the  number  of  incites  by  12. 

II.  SJtould  the  resulting  decimal  be  an  unending  one,  and 
it  is  desired  to  terminate  the  division  at  some  point,  say  the 
fourtJi  decimal  place,  carry  the  division  one  place  further, 
and  if  the  fifth  figure  is  o  or  greater  increase  the  fourth 
figure  by  1,  omitting  the  signs  -\-  and  — . 


EXAMPLES  FOR  PRACTICE. 

169.      Reduce  to  the  decimal  part  of  a  foot: 


(a)  3  in. 

(b)  4i  in. 

(c)  5  in.  Ans. 

(d)  6|in. 

(e)  11  in. 


(a)  .25  ft. 

(b)  .375ft. 

(c)  .4167  ft. 

(d)  .5521  ft. 

(e)  .9167  ft. 


TO  REDUCE  A  DECIMAL,  TO  A  FRACTION. 

170.  EXAMPLE. — Reduce  .125  to  a  fraction. 
SOLUTION.-    .125  =  ^  =  ft  =  J.     Ans. 
EXAMPLE. — Reduce  .875  to  a  fraction. 
SOLUTION.-    .875  =  fffo  =  fft  =  f.     Ans. 

171.  Rule. —  Under  the  figures  of  the  decimal,  place  1 
with  as  many  ciphers  at  its  right  as  there  are  decimal  places 
in  the  decimal,  and  reduce  the  resulting  fraction  to  its  lowest 
terms  by  dividing  both  numerator  and  denominator  by  the 
same  number. 


48 


ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 


172.      Reduce  the  following  to  common  fractions: 


(a) 
<*) 

(0 

(d) 
(*) 


(h) 


.125. 

.625. 

.3125. 

.04. 

.06. 

.75. 

.15625. 

.875. 


Ans. 


(«)  i- 

(*)  I- 

(0  A 

(<0  A- 

W  A- 


173.  To  express  a  decimal  approximatively  as  a 
fraction  having  a  given  denominator : 

174.  EXAMPLE. — Express  .5827  in  64ths. 

07  9Q2H 
SOLUTION.—      .5827  X  H  =       ^    •  Sa7  H- 

Hence,  .5827  =  f|,  nearly.     Ans. 

EXAMPLE. — Express  .3917  in  12ths. 

A  7004 

SOLUTION.—      .3917  X  if  =  -     r~.  say  A- 

1<& 

Hence,  .3917  =  T\,  nearly.     Ans. 

175.  Rule. — Reduce  1  to  a  fraction  having  the  given 
denominator.     Multiply  the  given  decimal  by  the  fraction  so 
obtained,  and  the  result  will  be  the  fraction  required. 


176. 


EXAMPLES  FOR  PRACTICE. 

Express : 

(a)     .625  in  8ths.  j-  (a) 

(*)      .3125  in  16ths. 
(c)      .15625  in  32ds. 


(d) 
(e) 

(/) 


.77  in  64ths. 
.81  in  48ths. 
.923  in  96ths. 


Ans. 


w 
(d) 


iV 

A- 


177.  The  sign  for  dollars  is  $.  It  is  read  dollars.  $25 
is  read  25  dollars. 

Since  there  are  100  cents  in  a  dollar,  1  cent  is  1  one-hun- 
dredth of  a  dollar;  the  first  two  figures  of  a  decimal  part  of 


§  1  ARITHMETIC.  49 

a  dollar  represent  cents.      Since  a  mill  is  ^  of  a  cent,  or 
"nmr  of  a  dollar,  the  third  figure  represents  mills. 

Thus,  $2-5.16  is  read  twenty-five  dollars  and  sixteen  cents; 
$25.168  is  read  twenty-Jive  dollars  sixteen  cents  and  ciglit 
mills. 


SYMBOLS  OF  AGGREGATION. 

178.  The  vinculum  —  — ,  parenthesis  ( ),  brackets  [  ], 
and  brace  {  \  are  called  symbols  of"  aggregation,  and  are 
used  to  include  numbers  which  are  to  be  considered  together; 
thus,  13x8  —  3,  or  13  X  (8  —  3),  shows  that  3  is  to  be  taken 
from  8  before  multiplying  by  13. 

13x(8-3)  =   13X5  =  65. 
13  X  8^3  =  13x5  =  65. 

When  the  vinculum  or  parenthesis  is  not  used,  we  have 
13x8-3  =  104-3  =  101. 

179.  In  any  series  of  numbers  connected  by  the  signs 
-4-,  — ,   X,   and  -=-,   the  operations  indicated  by    the  signs 
must  be  performed  in  order  from  left  to  right,  except  that 
no  addition  or  subtraction  may  be  performed  if  a  sign  of 
multiplication  or  division  follows  the  number  on  the  riglit 
of  a  sign  of   addition    or   subtraction    until    the   indicated 
multiplication  or  division  has  been  performed.      In  all  cases 
the  sign  of  multiplication  takes  the  precedence,  the  reason 
being  that  when  two  or  more  numbers  or  expressions  are 
connected  by  the  sign  of  multiplication  the  numbers  thus 
connected  are  regarded  as  factors  of  the  product  indicated, 
and  not  as  separate  numbers. 

EXAMPLE.— What  is  the  value  of  4  X  24  —  8  +  17  ? 
SOLUTION. — Performing  the  operations  in  order  from  left  to  right, 
4X24  =  96;  96-8  =  88;  88  +  17  =  105.     Ans. 

180.  EXAMPLE. — What  is  the  value  of  the  following  expression: 
1,296  -H  12  +  160  -  22  X  3£  =  ? 

SOLUTION.—  1,296-^-12  =  108;  108  +  160  =  268;  here  we  cannot  sub- 
tract 22  from  268  because  the  sign  of  multiplication  follows  22 ;  hence, 
multiplying  22  by  3£,  we  get  77,  and  268  —  77  =  191.  Ans. 


50  ARITHMETIC.  §  1 

Had  the  above  expression  been  written  1,296-^  12 +  160 
—  22x3^-^7  +  25,  it  would  have  been  necessary  to  have 
divided  22  X  3£  by  7  before  subtracting,  and  the  final  result 
would  have  been  22  X  3|  =  77;  77-5-7  =  11;  268-11  =  257; 
257  +  25  =  282.  Ans.  In  other  words,  it  is  necessary  to 
perform  all  the  indicated  multiplication  or  division  included 
between  the  signs  +  and  — ,  or  —  and  + ,  before  adding  or 
subtracting.  Also,  had  the  expression  been  written  1,296 
-i- 12  +  160  —  24£  -4-  7  X  3^  +  25,  it  would  have  been  necessary 
to  have  multiplied  3£  by  7  before  dividing  24^,  since  the 
sign  of  multiplication  takes  the  precedence,  and  the  final 
result  would  have  been  3|  X  7  =  24£ ;  24£  -=-  24£  =  1 ;  268 
-1  =267;  267  +  25  =  292.  Ans. 

It  likewise  follows  that  if  a  succession  of  multiplication 
and  division  signs  occur,  the  indicated  operations  must  not 
be  performed  in  order,  from  left  to  right — the  multiplication 
must  be  performed  first.  Thus,  24x3-7-4x2^-9x5  =  £. 
Ans.  In  order  to  obtain  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right,  symbols  of  aggregation  must  be  used. 
Thus,  by  using  two  vinculums  the  last  expression  becomes 
24x3-f-4x2-f-9x5  =  20,  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right. 


EXAMPLES  FOR  PRACTICE. 

181.      Find  the  values  of  the  following  expressions : 


(a) 

(6)  5  X  24  -  32. 

(0  5  X  24  H-  15. 

(d)  144  -  5  X  24. 

(e)  (1 , 691  -  540  +  559)  -*-  3  X  57. 
(/)  2,080  +  120  -  80  X  4  -  1,670. 
(#)  (90  +  60  -f-  25)  X  5  -  29. 

(K)  90  +  60  -r-  25  X  5. 


Ans. 


(a)  3. 

(b)  88. 

(c)  8. 

(d)  24. 

0?)  10. 

(/)  210. 

[(*)  1-2. 


ARITHMETIC. 

(CONTINUED.) 


PERCENTAGE. 

1.  Percentage   is  the  process  of  calculating-  by  Jntn- 
dredths. 

2.  The  term  per  cent,  is  an  abbreviation  of  the  Latin 
words  per  centum,  which  mean  by  the  hundred.      A  certain 
per  cent,  of  a  number  is  the  number  of  hundredths  of  that 
number  which  is  indicated  by  the  number  of  units  in  the 
percent.     Thus,  G  per  cent,   of  125  is  125XT£¥  =  7.5;  25 
per  cent,  of  80  is  80  X  TV7  =  20 ;  43  per  cent,  of  432  pounds  is 
432XT4</V  =  185. 7G  pounds. 

3.  The  sign   of  per  cent,    is   $,    and   is  read  per  cent. 
Thus,  6$  is  read  six  per  cent.;  12  L$  is  read  twelve  and  one- 
half  per  cent.,  etc. 

When  expressing-  the  per  cent,  of  a  number  to  iise  in  cal- 
culations, it  is  customary  to  express  it  decimally  instead  of 
fractionally.  Thus,  instead  of  expressing-  6$,  25$,  and  43$ 
as  T|^,  y2^,  and  TYtf,  it  is  usual  to  express  them  as  .00,  .25, 
and  .43. 

The  following-  table  will  show  how  any  per  cent,  can  be 
expressed  either  as  a  decimal  or  as  a  fraction : 


Per  Cent. 

Decimal. 

Fraction. 

Per  Cent. 

Decimal. 

Fraction. 

1£          

01 

, 

150$ 

1.50 

1  50   QJ.    t  1 

2*  

.02 

2    or  -p1^ 

500;?:  

5.00 

inn  Of  5 

5*.     . 

.05 

R       or     1 

i#.  . 

.0025 

\*  or  Ti,T 

10# 

.10 

10   or  JK 

4£  • 

.005 

,i_  or  TS'  „ 

25#  

25 

25     or     1 

\\%  

.015 

1  0  ff          200 
1J      ,,r       3 

50£ 

.50 

50   or    * 

84*.. 

.081 

SL  or  ,'• 

75*  

.75 

76     or     3 

12K.  . 

.125 

-^or    1 

100#  

1.00 

}SS  or  1 

161*.. 

.161 

'«?  or    ' 

125<£  

1.25 

4SS  or  H 

624*.  . 

.625 

t>2i   nr     r, 

2  ARITHMETIC.  §  2 

4.  The  names  of  the  different  elements  used  in  percent- 
age are :  the  base,  the  rate  per  cent. ,   the  percentage,   the 
amount,  and  the  difference. 

5.  The  base  is  the  number  on  which  the  per  cent,  is 
computed. 

6.  The  rate  is  the  number  of  hundredths  of  the  base  to 
be  taken. 

7.  The    percentage   is   the   part,   or  number    of   Jiun- 
dredtJis,  of  the  base  indicated  by  the  rate ;  or,  the  percentage 
is  the  result  obtained  by  multiplying  the  base  by  the  rate. 

Thus,  when  it  is  stated  that  7$  of  $25  is  $1.75,  $25  is  the 
base,  7$  is  the  rate,  and  $1.75  is  the  percentage. 

8.  The  amount  is  the  sum  of  the  base  and  percentage. 

9.  The  difference  is  the  remainder  obtained  by  sub- 
tracting the  percentage  from  the  base. 

Thus,  if  a  man  has  $180,  and  he  earns  6$  more,  he  will  have 
altogether  $180  +  $180 X. 06,  or  $180 +  $10. 80  =  $190.80. 
Here  $180* is  the  base;  6$,  the  rate;  $10.80,  the  percentage; 
and  $190.80,  the  amount. 

Again,  if  an  engine  of  125  horsepower  uses  16$  of  it  in 
overcoming  friction  and  other  resistances,  the  amount  left 
for  obtaining  useful  work  is  125  — 125  X.  16  =  125  —  20  =  105 
horsepower.  Here  125  is  the  base;  16$,  the  rate;  20,  the 
percentage ;  and  105,  the  difference. 

10.  From  the  foregoing  it  is  evident  that  to  find  the 
percentage,  the  base  must  be  multiplied  by  the  rate.      Hence, 
the  following 

Rule. —  To  find  the  percentage,  multiply  the  base  by  the 
rate  expressed  decimally. 

EXAMPLE. — Out  of  a  lot  of  300  bushels  of  apples  76$  were  sold.  How 
many  bushels  were  sold  ? 

SOLUTION. —  76$,  the  rate,  expressed  decimally,  is  .76;  the  base  is 
300 ;  hence,  the  number  of  bushels  sold,  or  the  percentage,  is,  by  the 
above  rule, 

300  X  .76  .=  228  bushels.     Ans. 

Expressing  the  rule  as  a 

Formula,  percentage  =  basey^rate. 


§  2  ARITHMETIC.  3 

11.  When  the  percentage  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  percentage  by  the  rate.      For, 
suppose  that  12  is  G$,  or  T£7,  of  some  number;  then  1$,  or 
yfj,  of  the  number,  is  12  -=-  G,  or  2.     Consequently,  if  2  =  \%, 
or  yfo,  100$,  or  ±™  =  2x  100  =  200.      But,  since  the  same 
result  may  be  arrived  at  by  dividing  12  by  .00,  for  12-=-.OG 
=  200,  it  follows  that : 

Rule. —  WJicn  tJic  percentage  and  rate  are  given,  to  find  tJie 
base,  divide  the  percentage  by  the  rate  expressed  decimally. 

Formula,  base  —  percentage -^  rate. 

EXAMPLE. — Bought  a  certain  number  of  bushels  of  apples  and  sold 
76$  of  them.  If  I  sold  228  bushels,  how  many  bushels  did  I  buy  ? 

SOLUTION. — Here  228  is  the  percentage,  and  76?",  or  .76,  is  the  rate; 
hence,  applying  the  rule, 

228  -i-  .76  =  800  bushels.     Ans. 

12.  When  the  base  and  percentage  are  given,  to  find  the 
rate,  the  rate  may  be  found,  expressed  decimally,  by  divi- 
ding the  percentage  by   the  base.      For,  suppose  that  it  is 
desired  to  find  what  per  cent.    12  is  of  200.      \%  of  200  is 
200 X. 01  =  2.      Now,  if  1$  is  2,  12  is  evidently  as  many  per 
cent,   as  the  number  of  times  that  2  is  contained  in  12,  or 
12  -r-  2  =  6$.      But  the    same  result   may  be   obtained  by 
dividing  12,  the  percentage,  by  200,  the  base,  since  12  -r-  200 
=  .06  =  6#.      Hence, 

Rule. —  When  the  percentage  and  base  are  given,  to  find 
the  rate,  divide  the  percentage  by  the  base,  and  the  result  will 
be  the  rate  expressed  decimally. 

Formula,  rate  —  percentage -v-'base. 

EXAMPLE. — Bought  300  bushels  of  apples  and  sold  228  bushels.  What 
per  cent,  of  the  total  number  of  bushels  was  sold  ? 

SOLUTION. — Here  300  is  the  base  and  228  is  the  percentage;  hence, 
applying  rule,  mte  =  ^  ^  m  =  76  =  ^  Ans 

EXAMPLE. — What  per  cent,  of  875  is  25  ? 

SOLUTION. — Here  875  is  the  base,  and  25  is  the  percentage;  hence, 
applying  rule,  gg  _^  8?g  =  >02f  =  2^ 

PROOF.—    875  X  .02$  =  25. 


ARITHMETIC.  §  2 


EXAMPLES    FOR    PRACTICE. 

13.      What  per  cent,  of: 
(a)     360  is  90  ? 


(t>)  900  is  360  ? 

(c)  125  is  25  ? 

(d)  150  is  750  ? 
(*)  280  is  112  ? 
(/)  400  is  200  ? 
(£•)  47  is  94  ? 
(h)  500  is  250  ? 


Ans. 


(£)  40*. 

(f)  20*. 

(</)  500*. 

(<•)  40*. 

(/)  50*. 

(£•)  200*. 


14.  The  amount  may  be  found,    when  the   base  and 
rate  are  given,  by  multiplying-  the  base  by  1  plus  the  rate, 
expressed  decimally.     For,  suppose  that  it  is  desired  to  find 
the  amount  when  200  is  the  base  and  6#  is  the  rate.     The 
percentage  is  200  X- 06  =  12,   and,  according  to  definition, 
Art.  8,  the  amount  is  200  +  12  =  212.     But,  the  same  result 
may  be  obtained  by  multiplying  200  by  l-f-.OG,  or   1.06, 
since  200X1.06  =  212.     Hence, 

~R\He.—  When  the  base  and  rate  are  given,  to  find  the 
amount,  multiply  the  base  by  1  plus  the  rate  expressed 
decimally. 

Formula,  amount  =.  basexQ -\-rate). 

EXAMPLE. — If  a  man  earned  $725  in  a  year,  and  the  next  year  10$ 
more,  how  much  did  he  earn  the  second  year  ? 

SOLUTION. — Here  725  is  the  base  and  10$  is  the  rate,  and  the  amount 
is  required.  Hence,  applying  the  rule, 

725  X  1.10  =  $797.50.     Ans. 

15.  When  the  base  and  rate  are  given,  the  difference 
may  be  found  by  multiplying  the  base  by  1  minus  the  rate 
expressed  decimally.     For,  suppose  that  it  is  desired  to  find 
the  difference  when  the  base  is  200  and  the  rate  is  6#.     The 
percentage  is  200 X. 06  =  12;  and,  according  to  definition, 
Art.   9,  the   difference  =  200  —  12  =  188.     But,  the   same 
result  may  be  obtained  by  multiplying  200  by  1  —  .06,  or  .94, 
since  200  X .  04  =  188.     Hence, 

Kule. —  When  the  base  and  rate  are  given,  to  find  the  differ- 
ence, multiply  the  base  by  1  minus  the  rate  expressed  decimally. 
Formula,  difference  =  base  X  (1  —  rate). 


§  2  ARITHMETIC.  5 

EXAMPLE.— Bought  300  bushels  of  apples  and  sold  all  but  24#  of 
them.  How  many  bushels  were  sold  ? 

SOLUTION. — Here  300  is  the  base,  24%  is  the  rate,  and  it  is  desired  to 
find  the  difference.  Hence,  applying  the  rule, 

300  X  (1  -  .24)  =  228  bushels.     Ans. 

16.  When  the  amount  and  rate  are  given,  the  base  may 
be  found  by  dividing  the  amount  by  1  plus  the  rate.      For, 
suppose   that   it   is  known    that   212   equals   some  number 
increased  by  G$  of  itself.     Then,  it  is  evident  that  212  equals 
106$  of  the  number  (base)  that  it  is  desired  to  find.     Conse- 

212 
quently,  if  212  -  10G$,  1$  =  —  =  2,  and  100$  -  2  X  100 

=  200  =  the  base.  But  the  same  result  may  be  obtained 
by  dividing  212  by  l  +  .OG,  or  l.OG,  since  212-=- LOG  =  200. 
Hence, 

Rule. —  When  the  amount  and  rate  are  given,  to  find  the 
base,  divide  the  amount  by  1  plus  the  rate  expressed  decimally. 
Formula,  base  =  amount-^-  (1  -\-rate). 

EXAMPLE. — The  theoretical  discharge  of  a  certain  pump  when  run- 
ning at  a  piston  speed  of  100  feet  per  minute  is  278,910  gallons  per  day 
of  10  hours.  Owing  to  leakage  and  other  defects,  this  value  is  25$ 
greater  than  the  actual  discharge.  What  is  the  actual  discharge  ? 

SOLUTION. — Here  278,910  equals  the  actual  discharge  (base)  increased 
by  25$  of  itself.  Consequently,  278,910  is  the  amount,  and  25#  is  the 
rate.  Applying  rule, 

actual  discharge  =  278,910--- 1.25  =  223,128  gallons.     Ans. 

17.  When  the  difference  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  difference  by  1  minus  the  rate. 
For,  suppose  that  188  equals  some  number  less  G$  of  itself. 
Then,  188  evidently  equals  100  —  G  =  94$  of  some  number. 
Consequently,  if  188  =  94$,  1$  =  188  -f-  94  =  2,   and  100$ 
=  2  x  100  =  200.     But  the  same  result  may  be  obtained  by 
dividing  188  by  1  — .  OG,  or  .  94,  since  188  -h .  94  =  200.   Hence, 

Rule. —  When  Hie  difference  and  rate  are  given,  to  Jind  the 
base,  divide  the  difference  by  1  minus  the  rate  expressed 
decimally. 

Formula,  base  =  difference -7- (\  —  rate). 

1-5 


6  ARITHMETIC.  §  2 

EXAMPLE. — Bought  a  certain  number  of  bushels  of  apples  and  sold 
76$  of  them.  If  there  were  72  bushels  left  unsold,  how  many  bushels 
did  I  buy  ? 

SOLUTION. — Here  72  is  the  difference  and  76$  is  the  rate.  Applying 
rule>  72  -T-  (1  -  .76)  =  300  bushels.  Ans. 

EXAMPLE. — The  theoretical  number  of  foot-pounds  of  work  per  min- 
ute required  to  operate  a  boiler  feed-pump  is  127,344.  If  30$  of  the 
total  number  actually  required  be  allowed  for  friction,  leakage,  etc., 
how  many  foot-pounds  are  actually  required  to  work  the  pump  ? 

SOLUTION. — Here  the  number  actually  required  is  the  base ;  hence, 
127,344  is  the  difference,  and  30$  is  the  rate.     Applying  the  rule, 
127,344  -r-  (1  -  .30)  =  181,920  foot-pounds.     Ans. 

18.  EXAMPLE. — A  certain  chimney  gives  a  draft  of  2.76  inches  of 
water.     By  increasing  the  height  20  feet,  the  draft  was  increased  to  3 
inches  of  water.     What  was  the  gain  per  cent.  ? 

SOLUTION. — Here  it  is  evident  that  3  inches  is  the  amount,  and  that 
2.76  inches  is  the  base.  Consequently,  3  —  2.76  =  .24  inch  is  the  per- 
centage, and  it  is  required  to  find  the  rate..  Hence,  applying  the  rule 
given  in  Art.  12, 

gain  per  cent.  =  .24  +  2.76  =  .087  =  8.7$.     Ans. 

19.  EXAMPLE. — A  certain  chimney  gave  a  draft  of  3  inches  of 
water.     After  an  economizer  had  been  put  in,  the  draft  was  reduced  to 
1.2  inches  of  water.     What  was  the  loss  per  cent.? 

SOLUTION. — Here  it  is  evident  that  1.2  inches  is  the  difference  (since 
it  equals  3  inches  diminished  by  a  certain  per  cent,  loss  of  itself),  and 
3  inches  is  the  base.  Consequently,  3  —  1.2  =  1.8  inches  is  the  percent- 
age. Hence,  applying  the  rule  given  in  Art.  12, 

loss  per  cent.  =1.8-f-3  =  .60  =  60$.     Ans. 

20.  To  ftml  the  gain  or  loss  per  cent.  : 

Rule. — Find  the  difference  between  the  initial  and  the  final 
value;  divide  this  difference  by  the  initial  value. 

EXAMPLE. — If  a  man  buys  a  house  for  $1,860,  and  some  time  after- 
wards builds  a  barn  for  25$  of  the  cost  of  the  house,  does  he  gain  or 
lose,  and  how  much  per  cent,  if  he  sells  both  house  and  barn  for 
§2,100? 

SOLUTION.— The  cost  of  the  barn  was  $1,860  X -25  =  $465;  conse- 
quently, the  initial  value,  or  total  cost,  was  $1,860 +  §465  =  §2,325. 
Since  he  sold  them  for  $2,100  he  lost  $2,325  — $2,100  =  $225.  Hence, 
applying  rule, 

225  H-  2,325  =  .0968  =  9.68$  loss.     Ans. 


ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

21.      Solve  the  following  : 

(a)  What  is  12^  of  $900  ?  f  (a)  $112.50. 

(b)  What  is  |£  of  627  ?  (/;)  5.016. 

(c)  What  is  331$  of  54  ?  (<r)  18. 

(d)  101  is  68f$  of  what  number  ?  A            (r/)  146 j£. 

(e)  784  is  83i$  of  what  number  ?  "   }   (e)  940.8. 


(/)     What  jj  of  960  is  160  ? 
Gr)     What  %  of  §3,606  is  $450f  ? 
(//)     What  %  of  280  is  112  ? 


K/0 


1.,  A  steam  plant  consumed  an  average  of  3,640  pounds  of  coal  per 
day.  The  engineer  made  certain  alterations  which  resulted  in  a  saving 
of  250  pounds  per  day.  What  was  the  per  cent,  of  coal  saved  ? 

Ans.  7^,  nearly. 

2.  If  the  speed  of  an  engine  running  at  126  revolutions  per  minute 
should  be  increased  6|^,  how  many  revolutions  per  minute  would  it 
then  make  ?  Ans.  134. 19  rev. 

3.  The  list  price  of  a  lot  of  silk  goods  is  $1,400,  of  some  laces  $1,150, 
and  of  some  calico  $340.     If  25/«  discount  was  allowed  on  the  silk,  22$ 
on  the  laces,  and  12|$  on  the  calico,  what  was  the  actual  cost  of  the 
purchase?  Ans.  $2,244.50. 

4.  If  I  loan  a  man  $1,100,  and  this  is  18£#  of  the  amount  that  I  have 
on  interest,  how  much  money  have  I  on  interest  ?  Ans.  $5,945.95. 

5.  A  test  showed  that  an  engine  developed  190.4  horsepower,  \~t% 
of  which  was  consumed  in  friction.     How  much  power  was  available 
for  use?  Ans.  161.84  H.  P. 

6.  By  adding  a  condenser  to  a  steam  engine,  the  power  was  increased 
14$ and  the  consumption  of  coal  per  horsepower  per  hour  was  decreased 
20#.     If  the  engine  could  originally  develop  50  horsepower,  and  required 
3£  pounds  of  coal  per  horsepower  per  hour,  what  would  be  the  total 
weight  of  coal  used  in  an  hour,  with  the  condenser,  assuming  the  engine 
to  run  full  power  ?  Ans.  159.6  pounds. 


DE^TOMI^ATE  NUMBERS. 

22.  A  denominate  number  is  a  concrete  number,  and 
may  be  either  simple  or  compound;  as,  8  quarts;  5  feet;  10 
inches,  etc. 

23.  A  simple  denominate  number  consists  of  units 
of  but  one  denomination;  as,  16  cents;   10  hours;  5  dollars, 
etc. 


8  ARITHMETIC.  §  2 

24.  A   compound  denominate  number   consists   of 
units  of  two  or  more  denominations  of  a  similar  kind ;  as,  3 
yards,  2  feet,  1  inch;  34  square  feet,  57  square  inches. 

25.  In  whole  numbers  and  in  decimals  the  law  of 

increase  and  decrease  is  on  the  scale  of  10,  but  in  com- 
pound or  denominate  numbers  the  scale  varies. 

26.  A  measure  is  a  standard  unit,  established  by  laiv 
or  custom,  by  which  quantity  of  any  kind  is  measured.     The 
standard  unit  of  dry  measure  is  the  Winchester  bushel ;  of 
weight,  the  pound ;  of  liquid  measure,  the  gallon,  etc. 

27.  Measures  are  of  six  kinds : 

1.  Extension.  4.     Time. 

2.  Weight.  5.     Angles. 

3.  Capacity.  6.     Money  or  value. 


MEASURES  OF  EXTENSION. 

28.     Measures   of  extension  are   used   in   measuring 
lengths,  distances,  surfaces,  and  solids. 


LINEAR  MEASURE. 

TABLE. 
Abbreviation. 


12    inches  (in.)  =  1  foot .    .     ft. 

3    feet  ...  =1  yard    .  yd. 

5.5  yards    .    .  =  1  rod  .    .    rd. 

40    rods  .    .    .  =  1  furlong  fur. 

8    furlongs   .  =  1  mile     .  mi. 


in.  ft.  yd.        rd.  fur.  mi. 

36  =    3  -  =     1 
198  =  16i     =5.5      =      1 
7,920  =  660     =220     =40  =  1 
63,360  =  5,280  =  1,760  =  320  =  8  =  1 


SURVEYOR'S  LINEAR  MEASURE. 

TABLE. 

7.92  inches  =  1  link li. 

25  links  =  1  rod rd. 

4  rods    ) 
100  links   j-  lchain     '     •     '     •     ch- 

80  chains  =  1  mile mi. 

mi.         ch.  rd.  li.  in. 

1    =    80    =    320    =    8,000    =    63,360 

29.     The  linear  unit,  generally  used  by  surveyors,  is 
Gunter's  chain,  which  is  equal  to  4  rods,  or  66  feet. 


ARITHMETIC. 


0 


3().  An  engineer's  chain,  iised  by  civil  engineers,  is 
100  feet  long,  and  consists  of  100  links.  In  computations, 
the  links  are  written  as  so  many  hundredths  of  a  chain. 

SQUARE  MEASURE. 

TABLE. 

144    square  inches  (sq.  in.)     .     .     .  = 

9    square  feet = 

30J  square  yards = 

160  square  rods - 

640  acres = 

sq.  mi.  A.  sq.  rd.  sq.  yd. 

1  -  640  =  102,400  =  3,097,600  =  27,878,400  =  4,014,489,600 


1  square  foot   . 

.     .     .     sq.  ft. 

1  square  yard  . 

.     .     .  sq.  yd. 

1  square  rod     . 

.     .     .    sq.  rd. 

1  acre 

....      A. 

1  square  mile  . 

.   sq.  mi. 

sq.  ft. 

sq.  in. 

SURVEYOR'S  SQUARE  MEASURE. 
TABLE. 

625  square  links  (sq.  li.)     .     .     .     .  =    1  square  rod     .     . 

16  square  rods =     1  square  chain 

10  square  chains =     1  acre       .     .     .     . 

640  acres =     1  square  mile  . 

36  square  miles  (6  mi.  square)  .     .   —     1  township  . 

sq.  mi.       A.        sq.  ch.       sq.  rd.  sq.  li. 

1      =  640  =  6,400  =  102,400  =  64,000,000 


sq.  rd. 
sq.  ch. 

.  A. 
sq.  mi. 

.   Tp. 


CUBIC!  MEASURE. 
TABLE. 

1,728    cubic  inches  (cu.  in.)    .     .     .  =     1  cubic  foot 

27    cubic  feet =     1  cubic  yard 

128    cubic  feet =     1  cord      .     . 

24f  cubic  feet =     1  perch    .     . 

cu.  yd.     en.  ft.       cu.  in. 
1       =    27    =    46,656 


cu.  ft. 

cu.  yd. 

.     cd. 

P. 


MEASURES  OF  WEIGHT. 

AVOIRDUPOIS  WEIGHT. 
TABLE. 

16  ounces  (oz.) =      1  pound  .     .     .     , 

100  pounds =    1  hundredweight 

20cwt,  or  2,000  Ib =     1  ton  .... 

T.        cwt.  Ib.  oz. 

1    =    20    =  2,000    =    32,000 


Ib. 

cwt. 

T. 


10  ARITHMETIC.  §  2 

31.  The  ounce  is  divided   into  halves,    quarters,    etc. 
Avoirdupois  weight  is  used  for  weighing  coarse  and  heavy 
articles.     One  avoirdupois  pound  contains  7,000  grains. 

LONG  TON  TABLE. 

16  ounces =     1  pound Ib. 

112  pounds =     1  hundredweight  .     .     .      cwt. 

20  cwt,  or  2,240  Ib =     1  ton T. 

32.  In  all  the  calculations  throughout  this  and  the  fol- 
lowing  sections,    2,000   pounds   will  be    considered   1   ton, 
unless  the  long  ton  (2,240  pounds)  is  especially  mentioned. 

TROY  WEIGHT. 

TABLE. 

24  grains  (gr.) =  1  pennyweight  ....  pwt. 

20  pennyweights =  1  ounce oz. 

12  ounces —  1  pound Ib. 

Ib.        oz.          pwt.  gr. 

1    =    12    =    240    =    5,760 

33.  Ti"oy  weight  is  used  in  weighing  gold  and  silver- 
ware, jewels,  etc.     It  is  used  by  jewelers. 


MEASURES  OF  CAPACITY. 
LIQUID  MEASURE. 

TABLE. 

4    gills  (gi.) =    1  pint pt. 

2    pints =     1  quart qt. 

4    quarts =    1  gallon gal. 

8H  gallons =    1  barrel bbl. 

2    barrels,  or  63  gallons  .     .     .     .  =    1  hogshead hhd. 

hhd.     bbl.       gal.         qt.  pt.             gi. 

1    =    2    =    63    =    252  =    504    =    2,016 

DRY  MEASURE. 

TABLE. 

2  pints  (pt.) =     1  quart „     .     qt. 

8  quarts =     1  peck pk. 

4  pecks =    1  bushel bu. 

bu.      pk.      qt.        pt. 
1   =  4  =  32  =  64 


ARITHMETIC. 


11 


MEASURE  OF  TIME. 

TABLE, 

60  seconds  (sec.) =  1  minute min. 

60  minutes .  =  1  hour hr. 

24  hours =  1  day da. 

7  days =  1  week wk. 

365  days        ) 

>• =     1  common  year    ....     yr. 

12  months  j 

366  days       ......          .     .  —     1  leap  year. 

100  years =     1  century. 

NOTE. — It  is  customary  to  consider  one  month  as  30  days. 

MEASURE  OF  ANGLES  OR  ARCS. 

TABLE. 

60  seconds  (") =     1  minute '. 

60  minutes =     1   degree °. 

90  degrees =     1  right  angle  or  quadrant  |    . 

360  degrees =     1  circle    .......  cir. 

1  cir.  =  360°  =  21,600'  =  1,296,000" 

MEASURE  OF  MONEY. 

UNITED    STATES    MONEY. 
TABLE. 

10  mills  (m.) =  1  cent ct. 

10  cents =  1  dime d. 

10  dimes =  1  dollar $. 

10  dollars =  1  eagle E. 

E.       §  d.  ct.  m. 

1  =  10  =  100  =  1,000  =  10,000 


MISCEUQANEOUS  TAIJL.E. 


12  things  are  1  dozen. 

12  dozen  are  1  gross. 

12  gross  are  1  great  gross. 

2  things  are  1  pair. 
20  things  are  1  score. 

1  league  is  3  miles. 

1  fathom  is  6  feet. 


1  meter  is  nearly  39.37  inches. 

1  hand  is  4  inches. 

1  palm  is  3  inches. 

1  span  is  9  inches. 
24  sheets  are  1  quire. 
20  quires,  or  480  sheets,  are  1  ream. 

1  bushel  contains  2,150.4  cubic  in. 
1  U.  S.  standard  gallon  (also  called  a  wine  gallon)  contains  231  cubic  in. 
1  U.  S.  standard  gallon  of  water  weighs  8.355  pounds,  nearly. 
1  cubic  foot  of  water  contains  7.481  U.  S.  standard  gallons,  nearly. 
1  British  imperial  gallon  weighs  10  pounds. 

It  will  be  of  great  advantage  to  the  student  to  carefully 
memorize  all  the  above  tables. 


12  ARITHMETIC.  §  2 

REDUCTION  OF  DENOMINATE  NUMBERS. 

34.  Reduction  of  denominate  numbers  is  the  process  of 
changing  their  denomination  without  changing  their  value. 
They  may  be  changed  from  a  higher  to  a  lower  denomina- 
tion, or  from  a  lower  to  a  higher — either  is  reduction.     As 

2  hours  =  120  minutes. 
32  ounces  =  2  pounds. 

35.  Principle. — Denominate  numbers  are  changed  to 
lower  denominations  by  multiplying,  and  to  higher  denom- 
inations by  dividing. 

To   reduce  denominate   numbers   to  lower  denom- 
inations : 

36.  EXAMPLE. — Reduce  5  yd.  2  ft.  7  in.  to  inches. 
SOLUTION. — 


yd.           ft. 
5             2 
3 
15ft. 
2ft. 
17ft. 
12 
34 
1  7 

in. 

7 

2  0  4  in. 
7  in. 

211  inches.     Ans. 

EXPLANATION. — Since  there  are  3  feet  in  1  yard,  in  5  yards 
there  are  5  X  3  or  15  feet,  and  15  feet  plus  2  feet  =  17  feet. 
There  are  12  inches  in  a  foot;  therefore,  12x17  =  204 
inches,  and  204  inches  plus  7  inches  =  211  inches  =  mim- 
ber  of  inches  in  5  yards  2  feet  and  7  inches. 

37.      EXAMPLE. — Reduce  6  hours  to  seconds. 
SOLUTION. —  6         hours. 

60 

360      minutes. 
60 


21600  seconds.     Ans. 


2 


ARITHMETIC. 


13 


EXPLANATION.: — As  there  are  GO  minutes  in  1  hour,  in  G 
hours  there  are  G  X  GO,  or  300,  minutes;  as  there  are  no  min- 
utes to  add,  we  multiply  3GO  minutes  by  GO,  to  get  the 
number  of  seconds. 

38.  In  order  to  avoid  mistakes,  if  any  denomination  be 
omitted,  represent  it  by  a  cipher.     Thus,  before  reducing  3 
rods  G  inches  to  inches,  insert  a  cipher  for  yards  and  a  cipher 
for  feet,  as 

rd.      yd.      ft.      in. 
3         0         0         G 

39.  Rule. — Multiply  the  number  representing  the  high- 
est denomination  by  the  number  of  units  in  the  ne.vt  lower 
required  to  make  one  of  the  higher  denomination,  and  to  the 
product  add  the  number  of  given  units  of  that  lower  denomi- 
nation.    Proceed  in  this  manner  until  the  number  is  reduced 
to  the  required  denomination. 


EXAMPLES  FOR  PRACTICE. 


4O.  Reduce:   ' 

(a)  4  rd.  2  yd.  2  ft.  to  ft. 

(b)  4  bu.  3  pk.  2  qt.  to  qt. 

(c)  13  rd.  5  yd.  2  ft.  to  ft. 

(d)  5  mi.  100  rd.  10  ft.  to  ft. 
(<?)  8  Ib.  4  oz.  6  pwt.  to  gr. 
(/)  52  hhd.  24  gal.  1  pt.  to  pt. 
(g)  5  cir.  16°  20'  to  minutes. 
(/i)  14  bu.  to  qt. 


Ans. 


74ft. 
154  qt. 
281.5  ft. 
28,000  ft. 
48,144  gr. 
26,401  pt. 
108,980'. 
448  qt. 


To  reduce  lower  to  higher  denominations : 

41.      EXAMPLE. — Reduce  211  inches  to  higher  denominations. 

SOLUTION.—  12 )  2  1  1  in. 

3)  1  7  ft. +7  in. 


5  yd.  +  2  ft.    Ans. 

EXPLANATION. — There  are  12  inches  in  1  foot;  therefore, 
211  divided  by  12  =  17  feet  and  7  inches  over.  There  are 
3  feet  in  1  yard;  therefore,  17  feet  divided  by  3  —  5  yards 


14  ARITHMETIC.  §  2 

and  2  feet  over.     The  last  quotient  and  the  two  remainders 
constitute  the  answer,  5  yards  2  feet  7  inches. 

42.      EXAMPLE. — Reduce    15,735    grains    Troy  weight   to    higher 
denominations. 

SOLUTION.—  2  4  )  1  5  7  3  5  gr.  (  6  5  5  pwt. 

144 
133 
1  20 


135 
120 


15gr. 

20)655  pwt.  (  3  2  oz. 
60 


40 

1  5  pwt. 

1  2  )  3  2  oz.  (  2  Ib. 
2_4 
8  oz. 

EXPLANATION. — There  are  24  grains  in  1  pennyweight,  and 
in  15,735  grains  there  are  as  many  pennyweights  as  24  is  con- 
tained in  15,735,  or  655  pennyweights  and  15  grains  remain- 
ing. There  are  20  pennyweights  in  1  ounce,  and  in  655 
pennyweights  there  are  32  ounces  and  15  pennyweights 
remaining.  There  are  12  ounces  in  1  pound,  and  in  32 
ounces  there  are  2  pounds  and  8  ounces  remaining.  The 
last  quotient  and  the  three  remainders  constitute  the  answer, 
2  pounds  8  ounces  15  pennyweights  15  grains. 

The  above  problem  is  worked  out  by  long  division,  because 
the  numbers  are  tpo  large  to  solve  easily  by  short  division. 
The  student  may  use  either  method. 

43.  Rule. — Divide  the  number  representing  the  denomi- 
nation given  by  the  number  of  units  of  this  denomination 
required  to  make  one  unit  of  the  next  higher  denomination. 
The  remainder  will  be  of  the  same  denomination,  but  the 
quotient  will  be  of  the  next  higher.  Divide  this  quotient  by 
the  number  of  units  of  its  denomination  required  to  make 
one  unit  of  the  next  higher.  Continue  until  the  highest 


ARITHMETIC. 


15 


denomination  is  readied,  or  until  tJiere  is  not  enough  of  a 
denomination  left  to  make  one  of  tJie  next  higher.  The  last 
quotient  and  the  remainders  constitute  the  required  result. 


EXAMPLES  FOTC  PRACTICE. 

44.      Reduce  to  units  of  higher  denominations: 

(a)  7,460  sq.  in.;  (f>)  7,580  sq.  yd.;  (c)  148,760  cu.  in.;  (it)  7,896 
cu.  ft.  toed.;  (e)  17,651";  (/)  1,120  cu.  ft.  to  cd.  ;  (g)  8,000  gi. ;  (//) 
36,450  Ib. 

(a)      5  sq.  yd.  6  sq.  ft.  116  sq.  in. 

(fi)      1  A.  90  sq.  rd.  17  sq.  yd.  4  sq.  ft.  72  sq.  in. 

(f)  3  cu.  yd.  5  cu.  ft.  152  cu.  in. 

(it)     61  cd.  88  cu.  ft. 
Ans. 

(e)      4   54  11. 

(/)     8  cd.  .96  cu.  ft. 

(g)  3  hhd.  61  gal. 

(//)      18  T.  4  cwt.  50  Ib. 


ADDITIOX  OF  DEXOMIXATE  NTTHBERS. 

45.      EXAMPLE.— Find  the  sum  of  3  cwt.  46  Ib.  12  oz.  ;   8  cwt.  12  Ib. 
13  oz. ;  12  cwt.  50  Ib.  13  oz.  ;  27  Ib.  4  oz. 
SOLUTION. — 


T. 

cwt. 

Ib. 

oz. 

0 

3 

46 

12 

0 

8 

12 

13 

0 

12 

50 

13 

0 

0 

27 

4 

1  4  37  10     Ans. 

EXPLANATION. — Begin  to  add  at  the  right-hand  column: 
4  +  13  +  134-12  =  42  ounces;  as  1C  ounces  make  1  pound, 
42  ounces  -=-16  =  2  and  a  remainder  of  10  ounces,  or  2 
pounds  and  10  ounces.  Place  10  ounces  under  ounce  column 
and  add  2  pounds  to  the  next  or  pound  column.  Then, 
2  +  27  +  50  +  12  +  46  =  137  pounds;  as  100  pounds  make 
a  hundredweight,  137  -f- 100  =  1  hundredweight  and  a 
remainder  of  37  pounds.  Place  the  37  under  the  pounds 
column,  and  add  1  hundredweight  to  the  next  or  hundred- 
weight column.  Next,  1  +  12  +  8  +  3  =  24  hundredweight. 


16  ARITHMETIC.  §  2 

20  hundredweight  make  a  ton;  therefore  24-^-20  =  1  ton 
and  4  hundredweight  remaining.  Hence,  the  sum  is  1  ton 
4  hundredweight  37  pounds  10  ounces.  Ans. 

46.      EXAMPLE.— What  is  the  sum  of  2  rd.  3  yd.  2  ft.  5  in. ;  6  rd.  1 

ft.  10  in. ;  17  rd.  11  in. ;  4  yd.  1  ft.? 

SOLUTION. — 


rd. 

yd. 

ft. 

in. 

2 

3 

2 

5 

6 

0 

1 

10 

17 

0 

0 

11 

0 

4 

1 

0 

26 

3^ 

0 

2 

26 

3 

1 

8 

Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  the  first 
column  =  2G  inches,  or  2  feet  and  2  inches  remaining.  The 
sum  of  the  numbers  in  the  next  column  plus  2  feet  =  6 
feet,  or  2  yards  and  0  feet  remaining.  The  sum  of  the  next 
column  plus  2  yards  =  9  yards,  or  9-j-5£  =  1  rod  and  3£ 
yards  remaining.  The  sum  of  the  next  column  plus  1  rod 
=  26  rods.  To  avoid  fractions  in  the  sum,  the  •£  yard  is 
reduced  to  1  foot  and  6  inches,  which  added  to  26  rods 
3  yards  0  feet  and  2  inches  =  26  rods  3  yards  1  foot  8 
inches.  Ans. 

47.  EXAMPLE.— What  is  the  sum  of  47  ft.  and  3  rd.  2  yd.  2  ft. 
10  in.? 

SOLUTION. — When  47  ft.  is  reduced  it  equals  2  rd.  4  yd.  2  ft.  which 
can  be  added  to  3  rd.  2  yd.  2  ft.  10  in.  Thus, 

rd.  yd.  ft.  in. 

3  2  2  10 

2 4 2 0 

6  li  1  10 

or  6  2  0  4    Ans. 

48.  Rule. — Place  the  numbers  so  that  like  denominations 
are  under  each  other.     Begin  at  the  right-hand  column,  and 
add.     Divide  the  sum  by  the  number  of  units  of  this  denomi- 
nation required  to  make  one  unit  of  the  next  higher.     Place 
the  remainder   under    the    column   added,    and   carry    the 
quotient  to  the  next  column.     Continue  in  this  manner  until 
the  highest  denomination  given  is  reached. 


§  2  ARITHMETIC.  17 

EXAMPLES  FOR  PRACTICE. 

49.      What  is  the  sum  of: 

(a)  25  Ib.  7  oz.  15  pwt.  23  gr.  ;  17  Ib.  16  pwt. ;  15  Ib.  4  oz.  12  pwt. ; 
18  Ib.  16  gr.  ;  10  Ib.  2  oz.  11  pwt.  16  gr.? 

(b)  9  mi.  13  rd.  4  yd.  2  ft.  ;  16  rd.  5  yd.  1  ft.  5  in. ;  16  mi.  2  rd.  3  in.  ; 
14  rd.  1  yd.  9  in.  ? 

(c)  3  cwt.  46  Ib.  12  oz.  ;  12  cwt.  9^  Ib. ;  2±  cwt.  21$  Ib.  ? 

(d)  10  yr.  8  mo.  5  wk.  3  da. ;  42  yr.  6  mo.  7  da. ;  7  yr.  5  mo.  18  wk. 
4  da. ;  17  yr.  17  da.  ? 

(e)  17  T.  11  cwt.  49  Ib.  14  oz. ;  16  T.  47  Ib.  13  oz. ;  20  T.  13  cwt.  14 
Ib.  6  oz. ;  11  T.  4  cwt.  16  Ib.  12  oz.? 

(/)  14  sq.  yd.  8  sq.  ft.  19  sq.  in. ;  105  sq.  yd.  16  sq.  ft.  240  sq.  in. ; 
43  sq.  yd.  28  sq.  ft.  165  sq.  in.? 

(a)  86  Ib.  3  oz.  16  pwt.  7  gr. 

(b)  25  mi.  47  rd.  1  ft.  5  in. 

(c)  18  cwt.  2  Ib.  14  oz. 

(d)  78  yr.  1  mo.  3  wk.  3  da. 

(e)  65  T.  9  cwt.  28  Ib.  13  oz. 
(/)  167  sq.  yd.  136  sq.  in. 


Ans. 


SUBTRACTION  OF  DENOMINATE  NUMBEKS. 

5O.  EXAMPLE.— From  21  rd.  2  yd.  2  ft.  6£  in.  take  9  rd.  4  yd. 
10i  in. 

SOLUTION. —               rd.  yd.  ft.  in. 

21  2  2  6} 

9  4  0  IQjr 

11  8J  1  8J    Ans. 

EXPLANATION. — Since  10^  inches  cannot  be  taken  from  G^ 
inches,  we  must  borrow  1  foot  or  12  inches  from  the  2  feet 
in  the  next  column  and  add  it  to  the  G|.  G|  +  12  —  18.;. 
18£  inches  — 10^  inches  =  8^  inches.  Then,  0  from  the  1 
remaining  foot  =  1  foot.  4  yards  cannot  be  taken  from  2 
yards;  therefore,  we  borrow  1  rod,  or  5£  yards,  from  21 
rods  and  add  it  to  2.  2  +  5£  =  7|;  7|-4  =  3J  yards. 
9  rods  from  20  rods  =  11  rods.  Hence,  the  remainder  is  11 
rods  3£  yards  1  foot  8£  inches.  Ans. 

To  avoid  fractions  as  much  as  possible,  we  reduce  the  \ 
yard  to  inches,  obtaining  18  inches;  this  added  to  8|  inches 
gives  26£  inches,  which  equals  2  feet  2^  inches.  Then,  2 
feet  -J- 1  foot  =  3  feet  =  1  yard,  and  3  yards  -f- 1  yard  =  4 


18  ARITHMETIC.  §  2 

yards.      Hence,  the  above  answer  becomes  11  rods  4  yards 
0  feet  2^  inches. 

51.  EXAMPLE. — What  is  the  difference  between  3  rd.  2yd.  2ft 
10  in.  and  47  ft.  ? 

SOLUTION.—    47  ft.  =  2  rd.  4  yd.  2  ft. 

rd.          yd.  ft.  in. 

3  2  2  10 

2420 

0  3£ 0~   ~10 

or  3  2  4     Ans. 

To  flnd  (approximately)  the  interval  of  time  between 
two  dates : 

52.  EXAMPLE. — How  many  years, months, days, and  hours  between 
4  o'clock  P.  M.  of  June  16, 1868,  and  10  o'clock  A.  M.,  September  29,  1891? 

SOLUTION. —  yr.         mo.         da.        hr. 

1891         8  28          10 

1868         5  15          16 

~23         3  12          18     Ans. 

EXPLANATION. — Counting  24  hours  in  1  day,  4  o'clock  p.  M. 
is  the  IGth  hour  from  the  beginning  of  the  day,  or  midnight. 
On  September  29,  8  months  and  28  days  have  elapsed,  and 
on  June  1C,  5  months  and  15  days.  After  placing  the  earlier 
date  under  the  later  date,  subtract  as  in  the  previous  prob- 
lems. Count  30  days  as  1  month. 

53.  Rule. — Place  the  smaller  quantity  under  the  larger 
quantity,  with  like  denominations  under  each  other.     Begin- 
ning at  the  right,  subtract  successively  the  number  in  the 
subtrahend  in  each  denomination  from  the  one  above,  and 
place   the  differences   underneath.     If  the  number   in   the 
minuend  of  any  denomination  is  less  than  the  number  under 
it  in  the  subtrahend,  one  must  be  borrowed  from  the  minuend 
of  the  next  higher  denomination,  reduced,  and  added  to  it. 


EXAMPLES  FOR  PRACTICE. 
54.      From: 

(a)   125  Ib.  8  oz.  14  pwt.  18  gr.  take  96  Ib.  9  oz.  10  pwt.  4  gr.  , 
(V)   126  hhd.  27  gal.  take  104  hhd.  14  gal.  1  qt.  1  pt. 

(c)  65  T.  14  cwt.  64  Ib.  10  oz.  take  16  T.  11  cwt.  14  oz. 

(d)  148  sq.  yd.  16  sq.  ft.  142  sq.  in.  take  132  sq.  yd.  136  sq.  in. 


2  ARITHMETIC.  19 

(e)   100  bu.  take  28  bu.  2  pk.  5  qt.  1  pt. 

(/)  14  mi.  34  rd.  16  yd.  13  ft.  11  in.  take  3  mi.  27  rd.  11  yd.  4  ft.  10  in. 

(a)  28  lb.  1 1  oz.  4  pwt.  14  gr. 

(b)  22  hhd.  12  gal.  2  qt.  1  pt. 

(c)  49  T.  3  cwt.  63  lb.  12  oz. 
Ans.  i  ^  ' 

(</)  16  sq.  yd.  16  sq.  ft.  6  sq.  in. 

(e)   71  bu.  1  pk.  2  qt.  1  pt. 
(/)  11  mi.  7  rd.  5  yd.  9  ft.  1  in. 


MULTIPLICATION  OF  DENOMINATE  NUMBERS. 

55.  EXAMPLE.—  Multiply  7  lb.  5  oz.  13  p\vt.  lo  gr.  by  12, 
SOLUTION. —  lb.          oz.         pwt.        gr. 

7  5  13  15 

12 

89  8~~         3        ~12    Ans. 

EXPLANATION.—  15  grains X  12  =  180  grains.  lSO-f-24 
=  7  pennyweights  and  12  grains  remaining.  Place  the  12 
in  the  grain  column  and  carry  the  7  pennyweights  to  the 
next.  Now,  13x12  +  7  =  163  pennyweights;  103-4-20  =  IS 
ounces  and  3  pennyweights  remaining.  Then,  5x12  +  8 
=  68  ounces;  08-^12  =  5  pounds  and  8  ounces  remaining. 
Then,  7x12  +  5  =  89  pounds.  The  entire  product  is  89 
pounds  8  ounces  3  pennyweights  12  grains.  Ans. 

56.  Rule. — Multiply     the    number     representing    cacJi 
denomination  by  the  multiplier  and  reduce  eacJi  product  to 
the  next  higher  denomination,  writing  the  remainders  under 
each  denomination,  and  carry  the  quotient  to  the  next,  as  in 
Addition  of  Denominate  Numbers. 

57.  In  multiplication  and  division  of  denominate  num- 
bers, it  is  sometimes  easier  to  reduce  the  number  to  the 
lowest  denomination  given  before  multiplying  or  dividing, 
especially  if  the  multiplier  or  divisor  is  a  decimal.     Thus, 
in  the  example  of  Art.  55,  had  the  multiplier  been  1.2,  the 
easiest  way  to  multiply  would  have  been  to  reduce  the  num- 
ber to  grains;  then,  multiply  by  1.2,  and  reduce  the  product 
to  higher  denominations.      For  example,  7  lb.  5  oz.  13  pwt. 
15  gr.  =  43,047  gr.     43,047x1.2  =  51,050.4  gr.  =  8  lb.  11 
oz.  12  pwt.  8.4  gr.     Also,  43,047x12  =  510,504 gr.   -.=  89  lb. 
8  oz.  3  pwt.  12  gr.,  as  above.     Either  method  may  be  used. 


20  ARITHMETIC. 


EXAMPLES  YOU  PRACTICE. 

58.      Multiply: 

(a)  15  cwt.  90  Ib.  by  5;  (6)  12  yr.  10  mo.  4  wk.  3  da.  by  14;  (c)  11  ml. 
145  rd.  by  20;  (d)  12  gal.  4  pt.  by  9;  (e)  8  cd.  76  cu.  ft.  by  15;  (/) 
4  hhd.  3  gal.  1  qt.  1  pt.  by  12. 

(a)  79  cwt.  50  Ib. 

(b)  180  yr.  11  mo.  2  wk. 


Ans. 


(c)  229  mi.  20  rd. 

(d)  112  gal.  2  qt. 

(e)  128  cd.  116  cu.  ft. 
(/)  48  hhd.  40  gal.  2  qt. 


DIVISION  OF  DENOMINATE  NUMBERS. 

59.      EXAMPLE.— Divide  48  Ib.  11  oz.  6  pwt.  by  8. 
SOLUTION. —  Ib.  oz.        pwt.        gr. 

8)48  11  6  0 

6  Ib.        1  oz.      8  pwt.    6  gr.     Ans. 

EXPLANATION. — After  placing  the  quantities  as  above, 
proceed  as  follows :  8  is  contained  in  48  six  times  without  a 
remainder.  8  is  contained  in  11  ounces  once,  with  3  ounces 
remaining.  3x20  =  60;  60  +  6  =  G6  pennyweights;  60 
penny  weights -7- 8  =  8  pennyweights  and  2  remaining; 
2x24  grains  =  48  grains;  48  grains -=-8  =  6  grains. 
Therefore,  the  entire  quotient  is  6  pounds  1  ounce  8  pen- 
nyweights 6  grains.  Ans. 

EXAMPLE. — A  silversmith  melted  up  2  Ib.  8  oz.  10  pwt.  of  silver,  which 
he  made  into  6  spoons ;  what  was  the  weight  of  each  spoon  ? 
SOLUTION. —  Ib.          oz.          pwt. 

6)2  8  10 

5  oz.         8  pwt.     8  gr.     Ans. 

EXPLANATION. — Since  we  cannot  divide  2  pounds  by  6,  we 
reduce  it  to  ounces.  2  pounds  =  24  ounces,  and  24  ounces 
-f-  8  ounces  =  32  ounces ;  32  ounces  -5-6  =  5  ounces  and  2 
ounces  over.  2  ounces  =  40  pennyweights ;  40  pennyweights 
+  10  pennyweights  =  50  pennyweights,  and  50  penny- 
weights -5-6  =  8  pennyweights  and  2  pennyweights  over.  2 
pennyweights  =  48  grains,  and  48  grains -5- 6  =  8  grains. 
Hence,  each  spoon  contains  5  ounces  8  pennyweights  8 
grains.  Ans. 


§2  ARITHMETIC. 

6O.      EXAMPLE.— Divide  820  rd.  4  yd.  2  ft.  by  112. 

rd.    yd.  ft.    rd.  yd.  ft.     in. 

SOLUTION.—       112)820    4    2  (    7     1     2     5. 143  Ans. 
784 

3  6  rd.  rem. 
5.5 


180 

l^JL 

F98.0yd. 

4 

1  1  2  )  2  0  2  yd.  (  1  yd. 
1  1  2 

9  0  yd.  rem. 
3 


2  7  0  ft. 
2ft. 


1  1  2  )  2  7  2  ft.  (  2  ft. 
224 

4  8  ft.  rem. 

12 

96 
48 

1  12)  5  7  Gin.  (5.1  4  2  8+ in.  or  5.143  in. 
560 

160 

112 


480 
448 
320 
224 
960 
896 
64 

EXPLANATION. — The  first  quotient  is  7  rods  with  3(5  rods 
remaining.  5.5x36  =  198 yards;  198 yards +  4 yards  =  202 
yards;  202  yards-i- 112  =  1  yard  and  90  yards  remaining. 
90x3  =  270 feet;  270 feet +  2 feet  =  272 feet;  272 feet -=-  112 
=  2  feet,  and  48  feet  remaining;  48x12  =  576  inches;  570 
inches -4- 112  =  5.143  inches,  nearly.  Ans. 

The  preceding  example  is  solved  by  long  division,  because 

1-6 


22  ARITHMETIC.  §  2 

the  numbers  are  too  large  to  deal  with  mentally.  Instead 
of  expressing-  the  last  result  as  a  decimal,  it  might  have 
been  expressed  as  a  common  fraction.  Thus,  576-5-112 
=  S^y  =  5|  inches.  The  chief  advantage  of  using  a  com- 
mon fraction  is  that  if  the  quotient  be  multiplied  by  the 
divisor,  the  result  will  always  be  the  same  as  the  original 
dividend. 

61.  Itule. — Find  how  many  times  the  divisor  is  contained 
in  the  first  or  highest  denomination  of '  l 'he  dividend.     Reduce 
the  remainder  (if  any]  to  the  next  lower  denomination,  and 
add  to  it  the  number  in  the  given  dividend  expressing  that 
denomination.     Divide  this   new   dividend  by   the   divisor. 
The  quotient  will  be  the  next  denomination  in  the  quotient 
required.      Continue  in  this  manner  until  the  lowest  denomi- 
nation is  reached.      The  successive  quotients  will  constitute 
the  entire  quotient. 

EXAMPLES  FOR  PRACTICE. 

62.  Divide: 

(a)  376  mi.  276  rd.  by  22;  (b)  1,137  bu.  3  pk.  4  qt.  1  pt.  by  10; 
(c)  84  cwt.  48  Ib.  49  oz.  by  16;  (<i)  78  sq.  yd.  18  sq.  ft.  41  sq.  in.  by  18; 
(*)  148  mi.  64  rd.  24  yd.  by  12;  (/)  100  T.  16  cwt.  18  Ib.  11  oz.  by  15; 
(g)  36  Ib.  18  oz.  18  pwt.  14  gr.  by  8;  (h)  112  mi.  48  rd.  by  100. 

(a)  17  mi.  41T7T  rd. 

(b)  113  bu.  3  pk.  1  qt.  \  pt. 

(c)  5  cwt.  28  Ib.  3TV  oz. 


Ans. 


(d)  4  sq.  yd.  4  sq.  ft.  2T55  sq.  in. 

(<?)  12  mi.  112  rd.  2  yd. 

(/)  6  T.  14  cwt.  41  Ib.  3*  *  oz. 

(g)  4  Ib.  8  oz.  7  pwt.  7f  gr. 

(A)  1  mi.  38||  rd. 


INVOLUTION. 

63.  Involution  is  the  process  of  multiplying  a  number 
by  itself  one  or  more  times.  The  product  obtained  by 
multiplying  a  number  by  itself  is  called  a  power  of  that 
number. 

Thus,  the  second  power  of  3  is  9,  since  3x3  are  9. 


§  2  ARITHMETIC.  23 

The  third  power  of  3  is  27,  since  3  X  3  X  3  are  27. 
The  fifth  power  of  2  is  32,  since  2x2x2X2X2  are  32. 

64.  An  exponent  is  a  small  figure  placed  to  the  right 
and  a  little  above  a  number  to  show  to  what/6>7ivr  it  is  to  be 
raised,  or  how  many  times  the  number  is  to  be  used  as  a 
factor,  as  the  small  figures  "•  3-  and  3  below. 

Thus,  3*  =  3X3  =  !). 

33  =  3X3X3  =  27. 

25  =  2X2X2X2X2  =  32. 

65.  The  root  of  a  number  is  that  number  which,  used 
the  required   number  of  times   as  a  factor,    produces  the 
number.      In  the  above  cases  3  is  a  root  of  J>,  since  3  X  3  are  9. 
It  is  also  a  root  of  27,  since  3x3x3  are  27.     Also,  2  is  a 
root  of  32,  since  2X2X2X2X2  are  32. 

66.  The  second    power    of    a    number    is    called    its 
square. 

Thus,  5'2  is  called  the  square  of  5,   or  5  squared,  and  its 

value  is  5  X  5  =  25. 

67.  The  third  power  of  a  number  is  called  its  cube. 
Thus,  53  is  called  the  cube  of  5,  or  5  cubed,  and  its  value 

is  5x5x5  =  125. 

To  fliitl  any  power  of  a  number : 

68.  EXAMPLE. — What  is  the  third  power,  or  cube,  of  35  ? 

SOLUTION.—  35  X  35  X  35, 

or  35 

35 


175 
105 

1225 
35 

6125 
3675 


cube  =  42875    Ans. 


ARITHMETIC. 


EXAMPLE. — What  is  the  fourth  power  of  15  ? 
SOLUTION.—  15  X  15  X  15  X  15, 

or  15 

L5. 

75 
15 
225 

15 


1125 

225 

3375 

15 

16875 
3375 
fourth  power  =  50625    Ans. 

69,      EXAMPLE.—    1.23  =  what? 
SOLUTION.—  1.2x1.2x1-2, 

or  1.2 

1.2 
1.44 


288 
144 
cube  =  1.728    Ans. 

7O.      EXAMPLE. — What  is  the  third  power,  or  cube,  of 

3X3X3 


SOLUTION. — 


Ans. 


"  8X8X8  " 

71.  Rule. — I.  To  raise  a  whole  number  or  a  decimal  to 
any  power ;  use  it  as  a  factor  as  many  times  as  there  are  units 
in  the  exponent. 

II.  To  raise  a  fraction  to  any  po^ver,  raise  both  the  numer- 
ator and  denominator  to  the  power  indicated  by  the  exponent. 


72. 


EXAMPLES  FOR  PRACTICE. 

Raise  the  following  to  the  powers  indicated : 


to 

(d) 

to 


85s. 

(I!)2- 
6.5s. 
14*. 
(t)3. 


Ans. 


</)    (I)3- 
(g)    (|)s. 

(ft)     1.45. 


(a) 
(*) 

to 

(^ 

w 


7,225. 

144 
Tff?' 

42.25. 
38,416. 


44. 

5.37824. 


§2  ARITHMETIC.  25 

EVOLFTIOX. 

73.  Evolution  is  the  reverse  of  involution.      It  is   the 
process  of  finding-  the  root  of  a  number  which  is  considered  as 
a  power. 

74.  The  square  root  of  a  number  is  that  number  which, 
when  used  twice  as  a  factor,  produces  the  number. 

Thus,  2  is  the  square  root  of  4,  since  2  X  2,  or  2*  =  4. 

75.  The  cube  root  of  a  number  is  that  number  which, 
when  used  three  times  as  a  factor,  produces  the  number. 

Thus,  3  is  the  cube  root  of  27,  since  3  X  3  X  3,  or  3s  —  27. 

76.  The  radical  sigfii  ^/,  when  placed  before  a  number, 
indicates  that  some  root  of  that  number  is  to  be  found. 

77.  The  index  of  the  root  is  a  small  figure  placed  over 
and  to  the  left  of  the  radical  sign,  to  show  what  root  is  to  be 
found. 

Thus,   v'lOO  denotes  the  square  root  of  100. 
4/125  denotes  the  cube  root  of  125. 
4/256  denotes  the  fourtJi  root  of  250,  and  so  on. 

78.  When  the  square  root  is  to  be  extracted,  the  index  is 
generally  omitted.     Thus,  vTIiO  indicates  the  square  root  of 
100.     Also,    1/225  indicates  the  square  root  of  225. 


SQUARE    ROOT. 

79.  The  largest  number  that  can  be  written  with  one 
figure  is  9,  and  92  =  81 ;  the  largest  number  that  can  be 
written  with  two  figures  is  99,  and  99"  =  9,801;  with  three 
figures  999,  and  999*  =  998,001;  with  four  figures  9,999,  and 
9,999s  =  99,980,001,  etc. 

In  each  of  the  above  it  will  be  noticed  that  the  square 
of  the  number  contains  just  twice  as  many  figures  as  the 
number. 

In  order  to  find  the  square  root  of  a  number,  the  first 
step  is  to  find  how  many  figures  there  will  be  in  the  root. 


26  ARITHMETIC.  §  2 

This  is  done  by  pointing  off  the  number  into  periods  of  two 
figures  each,  beginning  at  the  right.  The  number  of  periods 
will  indicate  the  number  of  figures  in  the  root. 

Thus,  the  square  root  of  83,740,801  must  contain  4  figures, 
since,  pointing  off  the  periods,  we  get  83'74'08'01,  or  4 
periods;  consequently,  there  must  be  4  figures  in  the  root. 
In  like  manner,  the  square  root  of  50,625  must  contain  3  fig- 
ures, since  there  are  (5'06'25)  3  periods. 

8O.      EXAMPLE.— Find  the  square  root  of  31,505,769. 

root 
SOLUTION.—        (a)         5  3  1'5  0'5  7'6  9  (  5  6  1  3     Aus. 

_5        (6)  2J> 
(d)  1  0  0       (c)   650 
6  636 

106       (e)     1457 
6  1121 


1120  33669 

1  33669 


1121  0 

1 

11220 

3 

11223 

EXPLANATION. — Pointing  off  into  periods  of  two  figures 
each,  it  is  seen  that  there  are  four  figures  in  the  root.  Now, 
find  the  largest  single  number  whose  square  is  less  than  or 
equal  to  31,  the  first  period.  This  is  evidently  5,  since  62  =  36, 
which  is  greater  than  31.  Write  it  to  the  right,  as  in  long 
division,  and  also  to  the  left  as  shown  at  (a).  This  is  the 
first  figure  of  the  root.  Now,  multiply  the  5  at  (a)  by  the  5 
in  the  root,  and  write  the  result  under  the  first  period,  as 
shown  at  (b].  Subtract,  and  obtain  6  as  a  remainder. 

Bring  down  the  next  period,  50,  and  annex  it  to  the 
remainder,  6,  as  shown  at  (e),  which  we  call  the  dividend. 
Add  the  root  already  found  to  the  5  at  (#),  getting  10,  and 
annex  a  cipher  to  this  10,  thus  making  it  100,  which  we  call 
the  trial  divisor.  Divide  the  dividend  (c)  by  the  trial 
divisor  (d]  and  obtain  6,  which  is  probably  the  next  figure 
of  the  root.  Write  6  in  the  root,  as  shown,  and  also  add  it 


§  2  ARITHMETIC.  27 

to  100,  the  trial  divisor,  making-  it  10G.  This  is  called 
the  complete  divisor. 

Multiply  this  by  G,  the  second  figure  in  the  root,  and  sub- 
tract the  result  from  the  dividend  (c).  The  remainder  is  14, 
to  which  annex  the  next  period,  making  it  1,457,  as  shown 
at  (e),  which  we  call  the  new  dividend.  Add  the  second 
figure  of  the  root  to  the  trial  divisor,  100,  and  annex  a 
cipher,  thus  getting  1,120.  Dividing  1,457  by  1,120,  we  get 
1  as  the  next  figure  of  the  root.  Adding  this  last  figure 
of  the  root  to  1,120,  multiplying  the  result  by  it,  and  sub- 
tracting from  1,457,  the  remainder  is  330. 

Annexing  the  next  and  last  period,  00,  the  result  is  33,000. 
Now,  adding  the  last  figure  of  the  root  to  1,121,  and  annex- 
ing a  cipher  as  before,  the  result  is  11,220.  Dividing  33,000 
by  11,220,  the  result  is  3,  the  fourth  figure  in  the  root. 
Adding  it  to  11,220  and  multiplying  the  sum  by  it,  the  result 
is  33,000.  .Subtracting,  there  is  no  remainder;  hence, 
4/31,505,700  =  5,013. 

81.  The  square  of  any  number  wholly  decimal  always 
contains  twice   as  many   figures  as  the   number   squared. 
For  example,  .I2  =  .01,   .13*  =  .0100,   .7512  =  .504001,  etc. 

82.  It  will  also  be  noticed  that  the  number  squared  is 
always  less  than  the  decimal.      Hence,  if  it  be  required  to 
find  the  square  root  of  a  decimal,  and  the  decimal  has  not  an 
even  number  of  figures  in  it,  annex  a  cipher.     The  best  way 
to  determine  the  number  of  figures  in  the  root  of  a  decimal 
is  to  begin  at  the  decimal  point,  and,   going  towards  the 
right,  point  off  the  decimal  into  periods  of  two  figures  each. 
Then,  if  the  last  period  contains  but  one  figure,  annex  a 
cipher. 

83.  EXAMPLE. — What  is  the  square  root  of  .000576  ? 

root 
SOLUTION.—  2  .0  O'O  5'7  6  ( .0  2  4    Ans. 

2_  4 

40  176 

__!  176 

44  0 


S8  ARITHMETIC.  §  2 

EXPLANATION. — Beginning  at  the  decimal  point,  and  point- 
ing off  the  number  into  periods  of  two  figures  each,  it  is  seen 
that  the  first  period  is  composed  of  ciphers ;  hence,  the  first 
figure  of  the  root  must  be  a  cipher.  The  remaining  portion 
of  the  solution  should  be  perfectly  clear  from  what  has 
preceded. 

84.  If  the  number  is  not  a  perfect  power,  the  root  will 
consist  of  an  interminable  number  of  decimal  places.     The 
result  may  be  carried  to  any  required  number  of  decimal 
places  by  annexing  periods   of   two   ciphers   each   to   the 
number. 

85.  EXAMPLE.— What  is  the  square  root  of  8  ?    Find  the  result  to 

five  decimal  places. 

root 

3.0  O'O  O'O  O'O  O'O  0  ( 1.7  3  2  0  5  +    Ans. 
1_ 

200 
189 
1100 
1029 
7100 
6924 
1760000 
1732025 
27975 


846400 

5 

346405 

EXPLANATION. — Annexing  five  periods  of  two  ciphers  each 
to  the  right  of  the  decimal  point,  the  first  figure  of  the  root 
is  1.  To  get  the  second  figure  we  find  that,  in  dividing  200 
by  20,  it  is  10.  This  is  evidently  too  large. 

Trying  9,  we  add  9  to  20,  and  multiply  29  by  9 ;  the  result 
is  261,  a  result  which  is  considerably  larger  than  200;  hence, 
9  is  too  large.  In  the  same  way  it  is  found  that  8  is  also  too 
large.  Trying  7,  7  times  27  are  189,  a  result  smaller  than 


§2  ARITHMETIC.  20 

200 ;  therefore,  7  is  the  second  figure  of  the  root.  The  next 
two  figures,  3  and  2,  are  easily  found.  The  fifth  figure  in 
the  root  is  a  cipher,  since  the  trial  divisor,  34,040,  is  greater 
than  the  new  dividend,  17,600.  In  a  case  of  this  kind  we 
annex  another  cipher  to  34,640,  thereby  making  it  346,400, 
and  bring  down  the  next  period,  making  the  17,600, 
1,760,000.  The  next  figure  of  the  root  is  5,  and,  as  we  now 
have  five  decimal  places,  we  will  stop. 

The  square  root  of    3    to    five   decimal  places   is,   then, 
1. 73205 +  . 

8G.      EXAMPLE. — What   is   the   square   root  of  .3   to  five  decimal 
places  ? 

root 

SOLUTION.—  5  .3  O'O  O'O  O'O  O'O  0  (  .5  4  7  7  2+  Ans. 

_5  2_5 

foO  500 

__4  41  6 

1 04  8400 

4  TJi  0  9 

1080  79100 

7  76629 

1087  247100 
219084 


10940  28016 

7 

10947 

7 

109540 

2 

109542 

EXPLANATION. — In  the  above  example  we  annex  a  cipher 
to  .3,  making-  the  first  period  .30,  since  every  period  of  a 
decimal,  as  was  mentioned  before,  must  have  two  figures 
in  it.  The  remainder  of  the  work  should  be  perfectly 
clear. 

87.  If  it  is  required  to  find  the  square  root  of  a  mixed 
number,  begin  at  the  decimal  point,  and  point  off  the  periods 
both  ways.  The  manner  of  finding  the  root  will  then  be 
exactly  the  same  as  in  the  previous  cases. 


30  ARITHMETIC. 

88.      EXAMPLE. — What  is  the  square  root  of  258.2449? 

root 
SOLUTION. — 


1 

2'58.24'49 

1 

1 

20 

158 

6 

156 

26 

2 

2449 

6 

2 

2449 

3200 

0 

7 

3207 

EXPLANATION. — In  the  above  example,  since  320  is  greater 
than  224,  we  place  a  cipher  for  the  third  figure  of  the  root, 
and  annex  a  cipher  to  320,  making  it  3,200.  Then,  bringing 
down  the  next  period,  49,  7  is  found  to  be  the  fourth  figure 
of  the  root.  Since  there  is  no  remainder,  the  square  root  of 
258.2449  is  1G.07. 

89.  Proof. —  To  prove   square   root,    square   the    result 
obtained.     "If  the  number  is  an  exact  power,   the  square  of 
the  root  will  equal  it;  if  it  is  not  an  exact  pozver,  the  square 
of  the  root  will  very  nearly  equal  it. 

90.  Rule. — I.     Begin  at  units  place,  and  separate  the 
number  into  periods  of  two  figures  each,  proceeding  from 
left  to  right  with  the  decimal  part,  if  there  be  any. 

II.  Find  the  greatest  number  whose  square  is  contained 
in  the  first,  or  left-hand,  period.      Write  this  number  as  the 
first  figure  in  the  root;  also,  write  it  at  the  left  of  the  given 
number. 

Multiply  this  number  at  the  left  by  the  first  figure  of  the 
root,  and  subtract  the  result  from  the  first  period;  then, 
annex  the  second  period  to  the  remainder. 

III.  Add  the  first  figure  of  the  root  to  the  number  in  the 
first  column  on  the  left,  and  annex  a  cipher  to  the  result; 
this  is  the  trial  divisor.     Divide  the  dividend  by  the  trial 
divisor  for  the  second  figure  in  the  root,  and  add  this  figure 
to  the  trial  divisor  to  form  the  complete  divisor.     Multiply 
the  complete  divisor  by  the  second  figure  in  the  root,  and 


§  2  ARITHMETIC.  31 

subtract  this  result  from  the  dividend.  (If  this  result  is  larger 
than  the  dividend,  a  smaller  number  must  be  tried  for  tin- 
second  figure  of  the  root. )  AW'  bring  down  the  third  period, 
and  annex  it  to  the  last  remainder  for  a  new  dividend.  Add 
the  second  figure  of  the  root  to  the  complete  divisor,  and  annex 
a  cipher  for  a  new  trial  divisor. 

IV.  Continue  in   this  manner  to  the  last  period,   after 
which,    if  any  additional  places  in   the  root  are   required, 
bringdown  cipher  periods,  and  continue  the  operation. 

V.  If  at  any  time  the  trial  divisor  is  not  contained  in  the 
dividend,  place  a  cipher  in  the  root,  annex  a  cipher  to  the 
trial  divisor,  and  bring  down  another  period. 

VI.  If  the  root  contains  an  interminable  decimal,  and  it 
is  desired  to  terminate  the  operation  at  some  point,  say,  the 
fourth  decimal  place,  carry  the  operation  one  place  further, 
and  if  the  fifth  figure  is  o  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  -f- . 

91.  Short  Method. — If  the  number  whose  root  is  to  be 
extracted  is  not  an  exact  square,  the  root  will  be  an  inter- 
minable decimal.  It  is  then  usual  to  extract  the  root  to  a 
certain  number  of  decimal  places.  In  such  cases,  the  work 
may  be  greatly  shortened  as  follows:  Determine  to  how 
many  decimal  places  the  work  is  to  be  carried,  say  f>,  for 
example;  add  to  this  the  number  of  places  in  the  integral 
part  of  the  root,  say  2,  for  example,  thus  determining  the 
number  of  figures  in  the  root,  in  this  case  5  +  2  =  7.  Divide 
this  number  by  2  and  take  the  next  higher  number.  In  the 
above  case,  we  have  7 -=-2  —  3^;  hence,  we  take  4,  the  next 
higher  number.  Now  extract  the  root  in  the  usual  manner 
until  the  same  number  of  figures  has  been  obtained  as  was 
expressed  by  the  number  obtained  above,  in  this  case  4. 
Then  form  the  trial  divisor  in  the  usual  manner,  but  omit- 
ting to  add  the  cipher;  divide  the  last  remainder  by  the 
trial  divisor  as  in  long  division,  obtaining  as  many  figures 
of  the  quotient  as  there  are  remaining  figures  of  the  root,  in 
this  case  7  —  4  =  3.  The  remainder  so  obtained  is  the 
remaining  figures  of  the  root. 


32  ARITHMETIC.  §  2 

Consider  the  example  in  Art.  86.  Here  there  are  5  fig- 
ures in  the  root.  We  therefore  extract  the  root  to  3  places 
in  the  usual  manner,  obtaining  .  547  for  the  first  three  root 
figures.  The  next  trial  divisor  is  1,094  (with  the  cipher 
omitted)  and  the  last  remainder  is  791.  Then,  791 -f- 1,094 
=  .723,  and  the  next  two  figures  of  the  root  are  72,  the 
whole  root  being  .  54772  -J-  .  Always  carry  the  division  one 
place  further  than  desired,  and  if  the  last  figure  is  5  or 
greater  increase  the  preceding  figure  by  1.  This  method 
should  not  be  used  unless  the  root  contains  five  or  more 
figures. 

If  the  last  figure  of  the  root  found  in  the  regular  man- 
ner is  a  cipher,  carry  the  process  one  place  further  before 
dividing  as  described  above. 


(a) 

(*) 

(c) 

(rtT) 

(e) 

(/) 

(g) 

(/t) 

(/) 

(f) 


EXAMPLES   FOR   PRACTICE. 

Find  the  square  root  of : 

186,624. 
2,050,624. 
29,855,296. 
..0116964. 
198. 1369. 
994,009. 

2.375  to  four  decimal  places. 
1.625  to  three  decimal  places. 
.3025. 
.571428. 
.78125. 


Ans.  - 


(a) 

432. 

0 

1,432. 

ft 

5,464. 

(d) 

.1081  + 

(e) 

14.0761. 

(f) 

997. 

(g) 

1.5411. 

(//) 

1.275. 

(') 

.55. 

Ul 

.7559  + 

(k) 

.8839. 

CUBE  ROOT. 

93.  In  the  same  manner  as  in  the  case  of  square  root, 
it  can  be  shown  that  the  periods  into  which  a  number 
is  divided,  whose  cube  root  is  to  be  extracted,  must  con- 
tain three  figures,  except  that  the  first,  or  left-hand,  period 
of  a  whole  or  mixed  number  may  contain  one,  two,  or  three 
figures. 


§  2  ARITHMETIC.  33 

94.      EXAMPLE.— What  is  the  cube  root  of  375,741,853,096  ? 
SOLUTION. — 

(!)        (3)  (3)          ,w, 

7       49          375'741'853'696(7216  Ans. 
J7        98          843 

14  14700  3274 1 

_7  434  30348 

210  15134  3493 853 

2  438  1  5  5  7301 

213  1555200         9  3649  30  90 

3  3161          936493896 

214  1557361  ~~0 
2         2163 

2160  155952300 
1        139816 

2161  156082116 
1 

2162 
1 

21630 
6 

21636 

EXPLANATION. — Write  the  work  in  three  columns,  as  fol- 
lows: On  the  right,  place  the  number  whose  cube  root  is  to 
be  extracted,  and  point  it  off  into  periods  of  three  figures 
each.  Call  this  column  (3).  Find  the  largest  number  whose 
cube  is  less  than  or  equal  to  the  first  period,  in  this  case  7. 
Write  the  7  on  the  right  as  shown,  for  the  first  figure  of  the 
root,  and  also  on  the  extreme  left  at  the  head  of  column  (1). 
Multiply  the  7  in  column  (1)  by  the  first  figure  of  the  root, 
7,  and  write  the  product,  49,  at  the  head  of  column  (2).  Mul- 
tiply the  number  in  column  (2)  by  the  first  figure  of  the 
root,  7,  and  write  the  product,  343,  under  the  figures  in  the 
first  period.  Subtract  and  bring  down  the  next  period, 
obtaining  32,741  for  the  dividend.  Add  the  first  figure 
of  the  root  to  the  number  in  column  (1),  obtaining  14, 
which  call  the  first  correction.  Multiply  the  first  correction 
by  the  first  figure  of  the  root,  add  the  product  to  the 
number  in  column  (2),  and  obtain  147.  Add  the  first  figure 
of  the  root  to  the  first  correction,  and  obtain  21,  which  call 


34  ARITHMETIC.  §  2 

the  second  correction.  Annex  two  ciphers  to  the  number 
in  column  (2),  and  obtain  14,700  for  the  trial  divisor;  also 
annex  one  cipher  to  the  second  correction,  and  obtain  210. 

Dividing  the  dividend  by  the  trial  divisor,  we  obtain  — /- — . 

14,700 

=  2  + ,  and  write  the  2  as  the  second  figure  of  the  root. 
Add  the  2  to  the  second  correction,  and  obtain  212,  which 
multiplied  by  the  second  figure  of  the  root  and  added  to 
the  trial  divisor,  gives  15,124,  the  complete  divisor.  This 
last  result  multiplied  by  the  second  figure  of  the  root  and 
subtracted  from  the  dividend,  gives  a  remainder  of  2,493. 
Annexing  the  third  period,  we  obtain  2,493,853  for  the  new 
dividend.  Adding  the  second  figure  of  the  root  to  the 
number  in  column  (1)  we  get  214  as  the  new  first  correction; 
this  multiplied  by  the  second  figure  of  the  root  and  added 
to  the  complete  divisor,  gives  15,552.  Adding  the  second 
figure  of  the  root  to  the  first  new  correction  gives  216  as  the 
second  new  correction.  Annexing  two  ciphers  to  the  num- 
ber in  column  (2)  gives  1,555,200,  the  new  trial  divisor. 
Annexing  one  cipher  to  the  second  new  correction  gives 
2,160.  Dividing  the  new  dividend  by  the  new  trial  divisor 

.    .      2,493,853 

we  obtain  -  =  1  -f- ,  and  wnte  1  as  the  third  figure 

1, 555, 200 

of  the  root.  The  remainder  of  the  work  should  be  perfectly 
clear  from  what  has  preceded. 

95.  In  extracting  the  cube  root  of  a  decimal,  proceed  as 
above,  taking  care  that  each  period  contains  three  figures. 
Begin  the  pointing  off  at  the  decimal  point,  going  towards 
the  right.     If  the  last  period  does  not  contain  three  figures, 
annex  ciphers  until  it  does. 

96.  EXAMPLE.— What  is  the  cube  root  of  .009129329  ? 

root 

SOLUTION.—  2  4  .0  0  9'1  2  9'8  2  9  ( .2  0  9 

2_        _8  8 

4          120000          1129329 
2  5481          1129329 


600       125481  0 

9 
609 


§  2  ARITHMETIC.  35 

EXPLANATION. — Beginning  at  the  decimal,  and  poin ting- 
off  as  shown,  the  largest  number  whose  cube  is  less  than  '.» 
is  seen  to  be  2 ;  hence,  2  is  the  first  figure  of  the  root. 
When  finding  the  second  figure,  it  is  seen  that  the  trial 
divisor,  1,200,  is  greater  than  the  dividend;  hence,  write  a 
cipher  for  the  second  figure  of  the  root ;  bring  down  the 
next  period  to  form  the  new  dividend ;  annex  two  ciphers 
to  the  trial  divisor  to  form  a  new  trial  divisor;  also,  annex 
one  cipher  to  the  (50  in  the  first  column.  Dividing  the  new 
dividend  by  the  new  trial  divisor,  we  get  Y^VoVif  =  '•'  +  > 
and  write  9  as  the  third  figure  of  the  root.  Complete  the 
work  as  before. 

97.      EXAMPLE.— What  is  the  cube  root  of  78,292.892952  ? 
SOLUTION. — 

root 

4  16  78'2  92.89  2'9  52  (4  2.7  8 

4          32  64 

8          4800          14292 
4  244  10088 


120         5044  4204892 

2  248  3766483 


122         529200  438409952 

2  8869  438409952 


124         538069 
2  8918 


1260        54698700 
7          102544 

1267        54801244 

7 

1274 

7 

12810 
.  8 


12818 


EXPLANATION. — Since  the  above  is  a  mixed  number,  begin 
at  the  decimal  point  and  point  off  periods  of  three  figures 


36  ARITHMETIC.  §  2 

each,  in  both  directions.  The  first  period  contains  but  two 
figures,  and  the  largest  number  whose  cube  is  less  than  78 
is  4 ;  consequently,  4  is  the  first  figure  of  the  root.  The 
remainder  of  the  work  should  be  perfectly  clear.  When 
dividing  the  dividend  by  the  trial  divisor  for  the  third  figure 
of  the  root,  the  quotient  was  8  + ,  but,  on  trying  it,  it  was 
found  that  8  was  too  large,  the  complete  divisor  being  con- 
siderably larger  than  the  trial  divisor.  Therefore,  7  was 
used  instead  of  8. 

O8.      EXAMPLE. — What  is  the  cube  root  of  5  to  five  decimal  places  ? 

SOLUTION. — 

root 


1 

1          5.0  0  O'O  0  O'O  0  O'O  0  O'O  0  0 

1 

2          1 

2 

300        4000 

1 

259        3913 

30 

559           87000000 

7 

308           78443829 

37 

8670000       8556171000 

7 

45981       7889992299 

44 

8715981        666178701000 

7 

46062        614014317973 

5100 

876204300      52164383027 

9 

461511 

5109 

876665811 

9 

461592 

5118 

87712740300 

9 

3590839 

51270 

87716331139 

9 

51279 

9 

51288 

9 

512970 

7 

512977 


§  2  ARITHMETIC.  37 

EXPLANATION. — In  the  above  example,  we  annex  five 
periods  of  ciphers,  of  three  ciphers  each,  to  the  5  for  the 
decimal  part  of  the  root,  placing  the  decimal  point  between 
the  5  and  the  first  cipher.  Since  it  is  easy  to  see  that  the 
next  figure  of  the  root  will  be  5,  we  increase  the  last  figure 
by  1,  obtaining  1.70998  for  the  correct  root  to  5  decimal 
places.  Ans. 

00.  EXAMPLE. — What  is  the  cube  root  of  .5  to  four  decimal 
places  ? 

SOLUTION. — 

root 

7  49  .500'000'000'000(.7937  + 

7  98  343 


14  14700  157000 

7  1971  150039 

210  16671           6961000 

9  2052           5638257 


219       1872300        1322743000 
9          7119        1321748953 


228      1879419  994047 

9          7128 


2370     188654700 
3         166579 


2373     188821279 
3 


2376 
3 

23790 

7 

23797 

EXPLANATION. — In  the  above  example,  we  annex  two 
ciphers  to  the  .5  to  complete  the  first  period,  and  three 
periods  of  three  ciphers  each.  The  cube  root  of  500  is  7 ; 
this  we  write  as  the  first  figure  of  the  root.  The  remainder 
of  the  work  should  be  perfectly  plain  from  the  explanations 
of  the  preceding  examples. 

1-7 


38  ARITHMETIC. 


1OO.      EXAMPLE. — What  is  the  cube  root  of  .05  to  four  decimal 


places  ? 
SOLUTION.  — 
3 
3 
6 
3 

9 
1  8 
2700 
576 

root 

.0  5  O'O  0  O'O  0  O'O  0  0  (.3  6  8  4  + 

27 

23000 
19656 

0 
2 

90 
6 

3276 
612 

334400 
318003 

96 

6 

388800 

8704 

16396 
16268 

8000 
5504 

1  02 
6 

397504 

8768 

1282496 

1080 
8 

40627200 
44176 

1088 
8 

40671376 

1096 

8__ 

1  1040 

_. 4 

1  1044 

101.  Proof. —  To  prove  cube   root,    cube   the  result  ob- 
tained.    If  the  given  number  is  an  exact  power,  the  cube  of 
the  root  will  equal  it;  if  not  an  exact  poiver,  the  cube  of  the 
root  will  very  nearly  equal  it. 

102.  Rule. — I.     Arrange   the  work   in  three  columns, 
placing  the  number  whose  cube  root  is  to  be  extracted,  in  the 
third,   or   right-hand,   column.     Begin  at   units  place,  and 
separate  the  number  into  periods  of  three  figures  each,  pro- 
ceeding from  the  decimal  point  towards  the  right  with  the 
decimal  part,  if  there  is  any. 

II.  Find  the  greatest  number  tv/iose  cube  is  not  greater 
than  the  number  in  the  first  period.  Write  this  number  as  the 
first  figure  of  the  root ;  also,  write  it  at  the  head  of  the  first 
column.  Multiply  the  number  in  the  first  column  by  the  first 
figure  in  the  root,  and  write  the  result  In  the  second  column. 
Multiply  the  number  in  the  second  column  by  the  first  figure 
of  the  root ;  subtract  the  product  from  the  first  period,  and 


§  2  ARITHMETIC.  39 

annex  the  second  period  to  the  remainder  for  a  new  dividend  ; 
add  the  first  figure  of  the  root  to  the  number  in  the  first 
column  for  I 'he  first  correction.  Multiply  the  first  correction 
by  the  first  figure  of  the  root,  and  add  the  product  to  the 
number  In  t lie  second  column.  Add  t lie  first  figure  of  the 
root  to  the  first  correction  to  form  the  second  correction. 
Annex  one  cipher  to  the  second  correction  and  two  ciphers  to 
the  last  number  in  the  second  column  ;  the  last  number  in  the 
second  column  is  the  trial  divisor. 

III.  Divide  the  dividend  by  the  trial  divisor  to  find  the 
second  figure  of  the  root.     Add  the  second  figure  of  the  root 
to  the  number  in  the  first  column,  multiply  the  sum  by  the 
second  figure  of  the  root,  and  add  the  result   to  the  trial 
divisor  to  form  the  complete  divisor,     Multiply  the  complete 
divisor  by  the  second  figure  of  the  root,  subtract  the  result 
from  the  dividend  in  the  third  column,  and  annex  the  third 
period  to  the  remainder  for  a  new  dividend.     Add  the  second 
figure  of  the  root  to  the  number  in  the  first  column  to  form 
the  first  correction  ;  multiply  t  lie  first  correction  by  the  second 
figure  of  the  root,  and  add  the  product  to  the  complete  divisor. 
Add  the  second  figure  of  the  root  to  the  first  correction  to 
form  the  second  correction.     Annex  one  cipher  to  the  second 
correction  and  two  ciphers  to  the  last  number  in  the  second 
column  to  form  the  new  trial  divisor. 

IV.  If  there  are  more  periods  to  be  brought  down,  proceed 
as  before.     If  there  is  a  remainder  after  the  root  of  the  last 
period  has  been  found,  annex  cipher  periods,  and  proceed  as 
before.     The  figures  of  the  root  thus  obtained  will  be  decimals. 

V.  If  the  root  contains  an  interminable  decimal,  and  it  is 
desired  to  terminate  the  operation  at   some  point,  say,   the 
fourth  decimal  place,  carry  the  operation  one  place  further , 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  -j- . 

1O3.  Art.  91  can  be  applied  to  cube  root  (or  any 
other  root)  as  well  as  to  square  root.  Thus,  in  the  exam- 
ple, Art.  98,  there  are  to  be  5  +  1  =  6  figures  in  the 
root.  Extracting  the  root  in  the  usual  manner  to  G  -r-  2  =  3, 


40  ARITHMETIC.  §  2 

say  4,  figures,  we  get  for  the  first  four  figures  1,709.  The 
last  remainder  is  8,556,171,  and  the  next  trial  divisor  with 
the  ciphers  omitted  is  8,762,043.  Hence,  the  next  two 
figures  of  the  root  are  8,556,171-^-8,762,043  =.976,  say  .98. 
Therefore,  the  root  is  1.70998. 


ROOTS  OF  FRACTIONS. 

104.  If  the  given  number  is  in  the  form  of  a  fraction, 
and  it  is  required  to  find  some  root  of  it,  the  simplest  and 
most  exact  method  is  to  reduce  the  fraction  to  a  decimal  and 
extract  the  required  root  of  the  decimal.     If,  however,  the 
numerator   and   denominator  of   the   fraction   are   perfect 
powers,  extract  the  required  root  of  each  separately,  and 
write  the  root  of  the  numerator  for  a  new  numerator,  and 
the  root  of  the  denominator  for  a  new  denominator. 

105.  EXAMPLE. — What  is  the  square  root  of  ^  ? 

/~9~         4/~9~ 
SOLUTION.—  I/TTT  =  -2-7=  =  I-     Ans. 


106.  EXAMPLE.  —  What  is  the  square  root  of  f  ? 
SOLUTION.—  Since  f  =  .625,    tf\  =    V^5=.7906.     Ans. 

107.  EXAMPLE.  —  What  is  the  cube  root  of       ? 


8/27 
SOLUTION.-  f  M  =  —  =  f.     Ans. 

108.  EXAMPLE.  —  What  is  the  cube  root  of  \  ? 
SOLUTION.—  Since  \  =.25,  tf\  =    ^25  =.62996  +  .     Ans. 

109.  Rule.  —  Extract  the  required  root  of  the  numer- 
ator and  denominator  separately  ;  or,  reduce  the  fraction  to 
a  decimal,  and  extract  the  root  of  the  decimal. 


EXAMPLES  FOR  PRACTICE. 

HO.      Find  the  cube  root  of: 

(<*)        /TV 

(b)      2  to  five  decimal  places. 


(c)      4,180,769,192.462  to  five  decimal  places. 


Ans. 


T- 


00 

(f)    513,229.783302144  to  three  decimal  places. 


00  I- 

(b)  1. 25992  +  . 

(c)  1,610.96238. 
\d)  .8862  +  . 
(e)  .7211  +  . 

[  (/)  80.064. 


§  2  ARITHMETIC.  41 

TO  EXTRACT    OTHER    HOOTS    THAN    THE  SQUARE 
AND  CUJJE  HOOTS. 

111.  ExAMi'LK.— What  is  the  fourth  root  of  256  ? 
SOLUTION. —  1/256  =  16. 

4/16  =  4. 
Therefore,  ^256  =  4.     Ans. 

In  this  example,  4/^5(5,  the  index  is  4,  which  equals  2  x2. 
The  root  indicated  by  2  is  the  square  root;  therefore,  the 
square  root  is  extracted  twice. 

112.  EXAMPLE.— What  is  the  sixth  root  of  64  ? 
SOLUTION. —  4/64  =  8. 

4'8  =  2. 
Therefore,  4  64  =  2.     Ans. 

In  this  example,  4/u'4,  the  index  is  0,  which  equals  2x3. 
The  root  indicated  by  3  is  the  cube  root;  therefore,  the 
square  and  cube  roots  are  extracted  in  succession. 

113.  Rule. — Separate  the  index  of  t lie  required  root  into 
its  factors  (2's  and  3's),  and  extract,  successively,  tlie  roots 
indicated  by  the  several  factors  obtained.      Tlie  final  result 
will  be  the  required  root. 

114.  EXAMPLE.— What  is  the  sixth   root  of  92,87:5,580  to    two 
decimal  places  ? 

SOLUTION. —  6  =  3x2.  Hence,  extract  the  cube  root,  and  tlic'ii 
extract  the  square  root  of  the  result.  4* '92,87:37580  =  452.8601,  and 
V35278601  =  21.28  +  .  Ans. 

115.  It  matters  not  which  root  is  extracted  first,  but  it 
is  probably  easier  and  more  exact  to  extract  the  cube  root 
first. 


EXAMPLES  FOR  PRACTICE. 
116.      Extract  the 

(a)  Fourth  root  of  100.  f  (a)    3.16227  +  . 

(b)  Fourth  root  of  3,049,800,625.  Ans.       (/>)     235. 

(c)  Sixth  root  of  9,474,296,896.  [  (f)     46. 


42  ARITHMETIC. 


RATIO. 

117.  Suppose  that  it  is  desired  to  compare  two  num- 
bers, say  20  and  4.     If  we  wish  to  know  how  many  times 
larger  20  is  than  4,  we  divide  20  by  4  and  obtain  5  for  the 
quotient;  thus,   20-^-4  =  5.      Hence,    we  say  that   20   is  5 
times  as  large  as  4,  i.  e. ,  20  contains  5  times  as  many  units 
as  4.     Again,  suppose  we  desire  to  know  what  part  of  20  is 
4.     We  then  divide  4  by  20  and  obtain  £;  thus,  4-4-20  =  |, 
or  .2.     Hence,  4  is  i  or  .2  of  20.     This  operation  of  com- 
paring two  numbers  is  termed  finding  the  ratio  of  the  two 
numbers.      Ratio,  then,  is  a  comparison.     It  is  evident  that 
the  two  numbers  to  be  compared  must  be  expressed  in  the 
same  unit ;  in  other  words,  the  two  numbers  must  both  be 
abstract  numbers  or  concrete  numbers  of  the  same  kind. 
For  example,  it  would  be  absurd  to  compare  20  horses  with 
4  birds,  or  20  horses  with  4.     Hence,  ratio  may  be  denned 
as  a  comparison  between  two  numbers  of  the  same  kind. 

118.  A  ratio  may  be  expressed  in  three  ways;  thus,  if 
it  is  desired  to  compare  20  and  4,  and  express  this  compari- 

20 

son  as  a  ratio,  it  may  be  done  as  follows :  20-4-4,  20  :  4,  or  — -. 

All  three  are  read  the  ratio  of  20  to  4.     The  ratio  of  4  to  20 

4 

would  be  expressed  thus :    4  -4-  20,   4  :  20,  or  — .     The  first 

/C\J 

method  of  expressing  a  ratio,  although  correct,  is  seldom  or 
never  used ;  the  second  form  is  the  one  of tenest  met  with, 
while  the  third  is  rapidly  growing  in  favor,  and  is  likely  to 
supersede  the  second.  The  third  form,  called  the  fractional 
form,  is  preferred  by  modern  mathematicians,  and  possesses 
great  advantages  to  students  of  algebra  and  of  higher  mathe- 
matical subjects.  The  second  form  seems  to  be  better 
adapted  to  arithmetical  subjects,  and  is  the  one  we  shall 
ordinarily  adopt.  There  is  still  another  way  of  expressing 
a  ratio,  though  seldom  or  never  used  in  the  case  of  a  simple 
ratio  like  that  given  above.  Instead  of  the  colon,  a  straight 
vertical  line  is  used ;  thus,  20  I  4. 


§  2  ARITHMETIC.  43 

110.  The  terms  of  a  ratio  are  the  two  numbers  to  be 
compared;  thus,  in  the  above  ratio,  20  and  4  are  the  terms. 
When  both  terms  are  considered  together  they  are  called  a 
couplet  ;  when  considered  separately,  the  first  term  is 
called  the  antecedent,  and  the  second  term,  the  conse- 
quent. Thus,  in  the  ratio  20  :  4,  20  and  4  form  a  couplet, 
and  20  is  the  antecedent,  and  4,  the  consequent. 

12O.  A  ratio  may  be  direct  or  invei-se.  The  direct 
ratio  of  20  to  4  is  20  :  4,  while  the  inverse  ratio  of  20  to  4  is 
4:20.  The  direct  ratio  of  4  to  20  is  4:20,  and  the  inverse 
ratio  is  20:4.  An  inverse  ratio  is  sometimes  called  a 
reciprocal  ratio.  The  reciprocal  of  a  number  is  1  divided 

by  the  number.     Thus,  the  reciprocal  of  17  is  —  ;   of   |  is 

1  < 

1-^f  =  f  ;    i.  e.  ,   the   reciprocal   of   a   fraction   is   the   frac- 
tion inverted.      Hence,  the  inverse  ratio  of  20  to  4  may  be 

expressed  as  4  :  20  or  as  ---  :    .     Both  have  equal  values  ;  for, 


121.  The  term  vary  implies  a  ratio.     When  we  say  that 
two  numbers  vary  as  some  other  two  numbers,  we  mean 
that  the  ratio  between  the  first  two  numbers  is  the  same  as 
the  ratio  between  the  other  two  numbers. 

122.  The  value  of  a  ratio  is  the  result  obtained  by  per- 
forming' the  division    indicated.      Thus,    the   value   of    the 
ratio  20  :  4  is  5  ;  it  is  the  quotient  obtained  by  dividing  the 
antecedent  by  the  consequent. 

123.  By  expressing  the  ratio  in  the  fractional  form,  for 

example,  the  ratio  of  20  to  4  as  —  ,  it  is  easy  to  see,   from 

4 

the  laws  of  fractions,  that  if  both  terms  be  multiplied  or 
both  divided  by  the  same  number  it  will  not  alter  the  value 
of  the  ratio.  Thus, 

20        20x5        100          .   20        20^4        5 

.  _     —  .      _      .—  _  •       o  *•*  rfH      .  _      .—  —       _     •  —  -      .  _ 

4      "    4X5      '    20  '  4      '    4-=-4    ""  1' 


44 


ARITHMETIC. 


124.  It  is  also  evident,  from  the  laws  of  fractions,  that 
multiplying  the  antecedent  or  dividing  the  consequent  mul- 
tiplies the  ratio,  and  dividing  the  antecedent  or  multiplying 
the  consequent  divides  the  ratio. 

125.  When  a  ratio  is  expressed  in  words,  as  the  ratio  of 
20  to  4,  the  first  number  named  is  always  regarded  as  the 
antecedent  and  the  second  as  the  consequent,  without  regard 
to  whether  the  ratio  itself  is  direct  or  inverse.      When  not 
otherwise  specified,  all  ratios  are   understood  to   be   direct. 
To  express  an  inverse  ratio  the  simplest  way  of  doing  it  is 
to  express  it  as  if  it  were  a  direct  ratio,  with  the  first  num- 
ber named  as  the  antecedent,  and  then  transpose  the  ante- 
cedent to  the  place  occupied  by  the  consequent  and  the 
consequent  to  the  place  occupied  by  the  antecedent;  or  if 
expressed  in  the  fractional  form,  invert  the  fraction.     Thus, 
to  express  the  inverse  ratio  of  20  to  4,  first  write  it  20 : 4, 

and  then,  transposing  the  terms,  as  4 : 20 ;  or  as  — ,  and 

4 

then  inverting,  as  — -.     Or,  the  reciprocals  of  the  numbers 


may  be  taken,  as  explained  above, 
transpose  its  terms. 


To  invert  a  ratio  is  to 


EXAMPLES  FOR  PRACTICE. 

126.     What  is  the  value  of  the  ratio  of: 
(a)      98  :  49  ? 
(£)      $45  :  §9  ? 


(<*) 

(/) 

(0 
C/) 
(*) 

3.5:4.5? 

The  inverse 
The  inverse 
The  inverse 
The  inverse 
The  ratio  of 
The  ratio  of 
The  ratio  of 
The  ratio  of 

ratio  of  76  to  19  ? 

ratio  of  49  to  98  ? 

ratio  of  18  to  24  ? 

ratio  of  9  to  15  ? 

10  to  3,  multiplied  by  3  ? 

85  to  49,  multiplied  by  7  ? 

18  to  64,  divided  by  9  ? 

14  to  28,  divided  by  5  ? 


Ans. 


(a) 
'(*) 

w 

(d) 

w 


(k) 

(0 

O) 


2. 
5. 
12*. 


10. 
5. 


127.     Instead  of  expressing  the  value  of  a  ratio  by  a 
single  number  as  above,  it  is  customary  to  express  it  by 


§2  ARITHMETIC.  45 

means  of  another  ratio  in  which  the  consequent  is  1.  Thus, 
suppose  that  it  is  desired  to  find  the  ratio  of  the  weights  of  two 
pieces  of  iron,  one  weighing  45  pounds  and  the  other  weigh- 
ing 30  pounds.  The  ratio  of  the  heavier  to  the  lighter  is 
then  45  :  30,  an  inconvenient  expression.  Using  the  frac- 

tional form,  we  have  ^—  •      Dividing  both  terms  by  .30,  the 

consequent,  we  obtain  -^  or  1|  :  1.  This  is  the  same  result 
as  obtained  above,  for  14-^1  =  14-,  and  45^30  =  1-1. 

128.  A  ratio  may  be  squared,  cubed,  or  raised  to  any 
power,  or  any  root  of  it  may  be  taken.  Thus,  if  the  ratio 
of  two  numbers  is  105  :  G3,  and  it  is  desired  to  cube  this 
ratio,  the  cube  may  be  expressed  as  1053  :  03s.  That  this  is 
correct  is  readily  seen  ;  for,  expressing  the  ratio  in  the  f  rac- 

.       .   -  105  .      .    /lor>\s       10.V 

tional  form,  it  becomes  —  -,  and  the  cube  is  I 

G-3  \  Oo  /          Go' 

=  105s  :  633.  Also,  if  it  is  desired  to  extract  the  cube  root 
of  the  ratio  1053  :  G33,  it  may  be  done  by  simply  dividing 
the  exponents  by  3,  obtaining  105  :  G3.  This  may  be  proved 
in  the  same  way  as  in  the  case  of  cubing  the  ratio.  Thus, 


(10^\3          /5\3 
^7rl    =UI>     i 
03  /          V3/ 


7rl    =UI>     it    follows    that    105s  :  033 


=  5s:  3s  (this  expression  is  read,  the  ratio  of  105  cubed 
to  63  cubed  equals  the  ratio  of  5  cubed  to  3  cubed),  and, 
hence,  that  the  antecedent  and  consequent  may  both  be 
multiplied  or  both  divided  by  the  same  number,  irrespec- 
tive of  any  indicated  powers  or  roots,  without  altering  the 
value  of  the  ratio.  Thus,  242:182  =  42:32.  For,  perform- 
ing the  operations  indicated  by  the  exponents,  24'  —  57<i 
and  18s  =  324.  Hence,  576:324  =  1|  or  1$:1.  Also,  4' 
=  16  and  3l  =  9;  hence,  16  :  9  =  1|  or  1£:1,  the  same 

result     as     before.       Also,    24'  ':  18?  =  j|i  = 

=  —  —  4*  •  32 
2  "        ' 


46  ARITHMETIC.  §  2 

The  statement  may  be  proved  for  roots  in  the  same 
manner.  Thus  1/W  :  &W  =  f  4s"  :  fF.  For,  the  #2& 
=  24  and  fls"3  =  18;  and,  24:18  =  1J  or  1£ :  1.  Also, 
Fg  =  4  and  f  3»  =  3 ;  4  :  3  =  1 J  or  1£  :  1. 

If  the  numbers  composing  the  antecedent  and  consequent 
have  different  exponents,  or  if  different  roots  of  those  num- 
bers are  indicated,  the  operations  above  described  cannot  be 
performed.  This  is  evident;  for,  consider  the  ratio  of 
42 :  8s.  When  expressed  in  the  fractional  form  it  becomes 

42  /4\2  /4\3 

--g,  which  cannot  be  expressed  either  as  I -I  or  as  I -1  ,  and, 

hence,  cannot  be  reduced  as  described  above. 


PROPORTION. 

13O.  Proportion  is  an  equality  of  ratios,  the  equality 
being"  indicated  by  the  double  colon  (  :  :  )  or  by  the  sign  of 
equality  (=}.  Thus,  to  write  in  the  form  of  a  proportion  the 
two  equal  ratios,  8  :  4  and  6  :  3,  which  both  have  the  same 
value,  2,  we  may  employ  one  of  the  three  following  forms  : 

8:4  ::  6:3         (1) 
8:4  =  6:3         (2) 


131.  The  first  form  is  the  one  most  extensively  used,  by 
reason  of  its  having  been  exclusively  employed  in  all  the 
older  works  on  mathematics.     The  second  and  third  forms 
are  being  adopted  by  all  modern  writers  on  mathematical 
subjects,  and,  in  time,  will  probably  entirely  supersede  the 
first  form.     In  this  subject  we  shall  adopt  the  second  form, 
unless  some  statement  can  be  made  clearer  by  using  the 
third  form. 

132.  A  proportion  may  be  read  in  two  ways.     The  old 
way  to  read  the  above  proportion  was  —  8  is  to  Jf.  as  6  is  to  3; 
the  new  way  is  —  the  ratio  of  8  to  4  equals  the  ratio  of  6  to  3. 
The  student  may  read  it  either  way,  but  we  Tecommend  the 
latter. 


§2  ARITHMETIC.  47 

133.  Each  ratio  of  a  proportion  is  termed  a  couplet. 
In  the  above  proportion,  8  :  4  is  a  couplet,  and  so  is  0  :  3. 

134.  The  numbers  forming-  the  proportion  are  called 
terms ;  and  they  are  numbered  consecutively  from  left  to 

light,   thus :  first   second     third  fourth 

8:4  =  6:3 

Hence,  in  any  proportion,  the  ratio  of  the  first  term  to 
the  second  term  equals  the  ratio  of  the  third  term  to  the 
fourth  term. 

135.  The  first  and  fourth  terms  of  a  proportion   are 
called  the  extremes,  and  the  second  and  third  terms,  the 
means.     Thus,  in  the  foregoing  proportion,  8  and  3  are  the 
extremes  and  4  and  6  are  the  means. 

136.  A  direct  proportion  is  one  in  which  both  coup- 
lets are  direct  ratios. 

137.  An  inverse  proportion  is  one  which  requires  one 
of  the  couplets  to  be  expressed  as  an  inverse  ratio.     Thus, 
8  is  to  4  inversely  as  3  is  to  (\  must  be  written  8:4  —  0:3; 
i.  e.,  the  second  ratio  (couplet)  must  be  inverted. 

138.  Proportion  forms  one  of  the  most  useful  sections 
of  arithmetic.    In  our  grandfathers'  arithmetics,  it  was  called 
"The  rule  of  three." 

139.  Rule  I. — /;/   any  proportion,    the  product  of  the 
extremes  equals  the  product  of  the  means. 

Thus,  in  the  proportion, 

17:51  =  14:42. 
17x42  =  51x14,  since  both  products  equal  714. 

140.  Rule  II. —  The  product  of  the  extremes  divided  by 
either  mean  gives  the  other  mean. 

EXAMPLE. — What  is  the  third  term  of  the  proportion  17  :  51  —     :  42  ? 
SOLUTION. — Applying  rule  1 1, 17  X  42  =  714,  and  714  -=-  51  =  14.  Ans. 

141.  Rule  III. —  The  product  of  the  means  divided  by 
either  extreme  gives  the  other  extreme. 

EXAMPLE. — What  is  the  first  term  of  the  proportion     :  51  —  14  :  42  ? 

SOLUTION.— Applying  rule   III,  51  X  14  =  714,   and  714-5-42  =  17. 

Ans. 


48  ARITHMETIC  §  2 

142.  When  stating-  a  proportion  in  which  one  of  the 
terms  is  unknown,  represent  the  missing-  term  by  a  letter, 
as  x.     Thus,  the  last  example  would  be  written, 

A-:  51  =  14:42 

51x14 

and  for  the  value  of  x  we  have  x  =  —  - —  =17. 

42 

143.  If  the  same  operations  (addition  and  subtraction 
excepted)  be  performed  upon  all  the  terms  of  a  proportion, 
the  proportion  is  not  thereby  destroyed.     In  other  words,  if 
all  the  terms  of  a  proportion  be  (1)  multiplied  or  (2)  divided 
by  the  same  number;  (3)  if  all  the  terms  be  raised  to  the 
same  power ;  (4)  if  the  same  root  of  all  the  terms  be  taken, 
or  (5)  if  both  couplets  be  inverted,  the  proportion  still  holds. 
We  will  prove  these  statements  by  a  numerical  example,  and 
the  student  can  satisfy  himself  by  other  similar  ones.     The 
fractional  form  will  be  used,  as  it  is  better  suited  to  the  pur- 
pose.   Consider  the  proportion  8:4  =  6:3.     Expressing  it  in 

8        6 
the  third  form,  it  becomes  —  =  —.     What  we  are  to  prove  is 

that  if  any  of  the  five  operations  enumerated  above  be 
performed  upon  all  the  terms  of  the  proportion,  the  first 
fraction  will  still  equal  the  second  fraction. 

8x7 

1.  Multiplying  all  the  terms  by  any  number,  say  7,  ^ — = 

6X7         56        42      AT       56  .   42 

~  3~X7'  °r  28  ~  21'  W  28  evldently  equals ^>  smce  the 

value  of  either  ratio  is  2,  and  the  same  is  true  of  the  original 
proportion. 

Q     .     fi 

2.  Dividing  all  the  terms  by  any  number,  say  7,       ] 

6-4-7         f       4  84  .63 

=  3^7' orT  =  I-     But?-?  =  2>  and  7-^-7  =  2also,the 

same  as  in  the  original  proportion. 

3.  Raising  all  the  terms  to  the  same  power,  say  the  cube, 

gS  Q3  QS  /S\3 

2j  =  03.     This  is  evidently  true,  since  75=  IT)  =  ^  —  8, 

3 
and       =          =  23  =  8  also. 


ARITHMETIC.  49 

4.      Extracting  the  same  root  of  all  the  terms,  say  the  cube 
_£S  _    jftj 

^4  "      ^3' 


A'TJ  ,3/77 

root,  —  =  =  —^=.      It  is  evident    that  this   is  likewise  true, 


f8 
since 


=  /!  = 


4        3 

5.     Inverting  both  couplets,  -  =  -,  which  is  true,  since 

o          u 

both  equal  •£. 

144.  If  both  terms  of  either  couplet  be  multiplied  or 
both  divided  by  the  same  number,   the  proportion  is  not 
destroyed.     This    should   be    evident    from    the   preceding 
article,  and  also  from  Art.  1*43.     Hence,  in  any  proportion, 
equal  factors  may  be  canceled  from  the  terms  of  a  couplet, 
before    applying   rule    II    or    III.       Thus,    the    proportion 
45  :  9  =  x\  7.  1,  we  may  divide  both  terms  of  the  first  coup- 
let by  9  (that  is,  cancel  9  from  both  terms),  obtaining  5  :  1 
=  x  :  7.  1,  whence  x  -  7.  1  X  5  -f-  1  =  35.  5.     (See  Art.  1  29.) 

145.  The  principle  of  all  calculations  in  proportion  is 
this  :   Three  of  the  terms  are  always  given,  and  the  remain- 
ing one  is  to  be  found. 

146.  EXAMPLE.  —  If  4  men  can  earn  §25  in  one  week,  how  much 
can  12  men  earn  in  the  same  time  ? 

SOLUTION.  —  The  required  term  must  bear  the  same  relation  to  the 
given  term  of  the  same  kind,  as  one  of  the  remaining  terms  bears  to 
the  other  remaining  term.  We  can  then  form  a  proportion  by  which 
the  required  term  may  be  found. 

The  first  question  the  student  must  ask  himself  in  every  calculation 
by  proportion  is  : 

"What  is  it  I  want  to  find  ?" 

In  this  case  it  is  dollars.  We  have  two  sets  of  men,  one  set  earning 
$25,  and  we  want  to  know  how  many  dollars  the  other  set  earns.  It  is 
evident  that  the  amount  12  men  earn  bears  the  same  relation  to  the 
amount  4  men  earn  as  12  men  bear  to  4  men.  Hence,  we  have  the 
proportion,  the  amount  12  men  earn  is  to  §25  as  12  men  are  to  4  men, 
or,  since  either  extreme  equals  the  product  of  the  means  divided  by 
the  other  extreme,  we  have 

The  amount  12  men  earn  :  $25  ::  12  men  :  4  men, 

$^5  V  12 
or  the  amount  12  men  earn  =  ~~A  —  ~  =  $75.     Ans. 


50  ARITHMETIC.  §  2 

Since  it  matters  not  which  place  .v,  or  the  required  term,  occupies, 
the  problem  could  be  stated  in  any  of  the  following  forms,  the  value  of 
x  being  the  same  in  each : 

(a)  §25  :  the  amount  12  men  earn  =  4  men  :  12  men ;  or  the  amount 

§25  X  12 
12  men  earn  =  ^ — -,  or  §75,  since  either  mean  equals  the  product 

of  the  extremes  divided  by  the  other  mean. 

(b)  4  men  :  12  men  =  §25  :  the  amount  that  12  men  earn ;  or  the 

825  V  12 

amount  that   12  men   earn  =  -. ,  or  §75,  since   either  extreme 

4 

equals  the  product  of  the  means  divided  by  the  other  extreme. 

(c)  12  men  :  4  men  =  the  amount  12  men  earn  :  §25;  or  the  amount 

§25  V  12 
that  12  men  earn  =  —  '—^ — -,  or  §75,    since   either  mean   equals  the 

product  of  the  extremes  divided  by  the  other  mean. 

147.  If  the  proportion  is  an  inverse  one,  first  form  it  as 
though  it  were  a  direct  proportion,  and  then  invert  one  of 
the  couplets. 

EXAMPLES  FOR  PRACTICE. 

Find  the  value  of  x  in  each  of  the  following: 


(a)  §16  :  §64  ::  x  :  §4. 

(b)  x  :  85  ::  10  :  17. 

(c)  24  :  x  ::  15  :  40. 

(d)  18:94::2:.r.  Ans.  - 
(<?)  §75  :  §100  =  x  :  100. 

(/)  15  pwt. :  x  =  21  :  10. 

(g)  x  :  75  yd.  =  $15  :  §5. 


(a)  x  =  *1. 

(b)  x  =  50. 

(c)  x  =  64. 
(rf)  x  =  lOf 
(e)  x  =  75. 

(/)  x  =  1\  pwt. 

I  (g)  x  =  225  yd. 


1.  If  75  pounds  of  lead  cost  §2.10,  what  would  125  pounds  cost  at 
the  same  rate  1  Ans.  §3.50. 

2.  If  A  does  a  piece  of  work  in  4  days  and  B  does  it  in  7  days,  how 
long  will  it  take  A  to  do  what  B  does  in  63  days  1  Ans.  36  days. 

3.  The  circumferences  of  any  two  circles  are  to  each  other  as  their 
diameters.     If  the  circumference  of  a  circle  7  inches  in  diameter  is  22 
inches,  what  will  be  the  circumference  of  a  circle  31  inches  in  diameter  ? 

Ans.  97f  inches. 

INVERSE    PROPORTION. 

149.  In  Art.  137,  an  inverse  proportion  was  denned  as 
one  which  required  one  of  the  couplets  to  be  expressed  as  an 
inverse  ratio.  Sometimes  the  word  inverse  occurs  in  the 
statement  of  the  example;  in  such  cases,  the  proportion  can 


§  2  ARITHMETIC.  51 

be  written  directly,  merely  inverting  one  of  the  couplets. 
But  it  frequently  happens  that  only  by  carefully  studying 
the  conditions  of  the  example,  can  it  be  ascertained  whether 
the  proportion  is  direct  or  inverse.  When  in  doubt,  the 
student  can  always  satisfy  himself  as  to  whether  the  propor- 
tion is  direct  or  inverse  by  first  ascertaining  what  is  required, 
and  stating  the  proportion  as  a  direct  proportion.  Then,  in 
order  that  the  proportion  may  be  true,  if  the  first  term  is 
smaller  than  the  second  term,  the  third  term  must  be  smaller 
than  the  fourth ;  or  if  the  first  term  is  larger  than  the  second 
term,  the  third  term  must  be  larger  than  the  fourth  term. 
Keeping  this  in  mind,  the  student  can  always  tell  whether 
the  required  term  will  be  larger  or  smaller  than  the  other 
term  of  the  couplet  to  which  the  required  term  belongs. 
Having  determined  this,  the  student  then  refers  to  the 
example,  and  ascertains  from  its  conditions  whether  the 
required  term  is  to  be  larger  or  smaller  than  the  other  term 
of  the  same  kind.  If  the  two  determinations  agree,  the  pro- 
portion is  direct,  otherwise  it  is  inverse,  and  one  of  the 
couplets  must  be  inverted. 

15O.  EXAMPLE. — If  A's  rate  of  doing  work  is  to  B's  as  5  :  7,  and  A 
does  a  piece  of  work  in  42  days,  in  what  time  will  B  do  it  ? 

SOLUTION. — The  required  term  is  the  number  of  days  it  will  take  B 
to  do  the  work.  Hence,  stating  as  a  direct  proportion, 

5  :  7  =  42  :  .r. 

Now,  since  7  is  greater  than  5,  .r  will  be  greater  than  42.  But,  refer- 
ring to  the  statement  of  the  example,  it  is  easy  to  see  that  B  works 
faster  than  A ;  hence  it  will  take  B  a  less  number  of  days  to  do  the 
work  than  A.  Therefore,  the  proportion  is  an  inverse  one,  and  should 

be  stated 

5  :  7  =  .r  :  42 

from  which  x  =  — ~- —  =  30  days.     Ans. 

Had  the  example  been  stated  thus:  The  time  that  A  requires  to  do 
a  piece  of  work  is  to  the  time  that  B  requires,  as  5  :  7 ;  A  can  do  it  in  42 
days,  in  what  time  can  B  do  it?  it  is  evident  that  it  would  take  B  a 
longer  time  to  do  the  work  than  it  would  A ;  hence,  .r  would  be  greater 

7  X  42 
than  42,  and  the  proportion  would  be  direct,  the  value  of  .r  being  — ^ — 

=  58.8  days. 


52  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

151.  Solve  the  following: 

1.  If  a  pump  which  discharges  4  gal.  of  water  per  min.  can  fill  a  tank 
in  20  hr.,  how  long  will  it  take  a  pump  discharging  12  gal.  per  min.  to 
fill  it?  Ans.  6|hr. 

2.  If  a  pump  discharges  90  gal.  of  water  in  20  hr.,  in  what  time  will 
it  discharge  144  gal.  ?  Ans.  32  hr. 

3.  The  weight  of  any  gas  (the  volume  and  pressure  remaining  the 
same)  varies  inversely  as  the  absolute   temperature.      If  a  certain 
quantity  of  some  gas  weighs  2.927  Ib.  when  the  absolute  temperature 
is  525°,  what  will  the  same  volume  of  gas  weigh  when  the  absolute 
temperature  is  600°,  the  pressure  remaining  the  same  ?    Ans.  2.561+  Ib. 

4.  If  50  cu.  ft.  of  air  weigh  4.2  pounds  when  the  absolute  tempera- 
ture is  562°,  what  will  be  the  absolute  temperature  when  the  same 
volume  weighs  5.8  pounds,  the  pressure  being  the  same  in  both  cases  ? 

Ans.  407°,  very  nearly. 

POWERS    AND    ROOTS    IN   PROPORTION. 

152.  It  was  stated  in  Art.  128  that  a  ratio  could  be 
raised  to  any  power  or  any  root  of  it  might  be  taken.     A 
proportion  is  frequently  stated  in  such  a  manner  that  one  of 
the  couplets  must  be  raised  to  some  power  or  some  root  of 
it  must  be  taken.     In   all   such  cases,   both  terms  of  the 
couplet  so  affected  must  be  raised  to  the  same  power  or  the 
same  root  of  both  terms  must  be  taken. 

153.  EXAMPLE.  —  Knowing  that  the  weight  of  a  sphere  varies  as 
the  cube  of  its  diameter,  what  is  the  weight  of  a  sphere  6  inches  in 
diameter  if  a  sphere  8  inches  in  diameter  of  the  same  material  weighs 
180  pounds  ? 

SOLUTION.  —  This  is  evidently  a  direct  proportion.     Hence,  we  write 

63  :  83  =  x  :  180. 

Dividing  both  terms  of  the  first  couplet  by  23  (see  Art.  129) 
3s  :  4s  =  x  :  180,  or  27  :  64  =  x  :  180; 

27X180 
whence,  x  =  —  ^  —  =  75}|  pounds.     Ans. 

EXAMPLE.  —  A  sphere  8  inches  in  diameter  weighs  180  pounds;  what 
is  the  diameter  of  another  sphere  of  the  same  material  which  weighs 
75^f  pounds  ? 

SOLUTION.  —  Since  the  weights  of  any  two  spheres  are  to  each  other 
as  the  cubes  of  their  diameters,  we  have  the  proportion 
180  : 


§  2  ARITHMETIC.  53 

The  required  term,  .1;  must  be  cubed,  because  the  other  term  of  the 
couplet  is  cubed  (see  Art.  152).     But,  y:!  =  513;  hence, 


180 


loU 
whence,  x  —    ^2W  =  6  inches.     Ans. 

154.  Since  taking-  the  same  root  of  all  the  terms  of  a 
proportion  does  not  change  its  value  (Art.  143),  the  above 
example  might  have  been  solved  by  extracting  the  cube  root 
of  all  the  numbers,  thus  obtaining-  ^180  : 


8  X     75| 

whence,  x  =  -  --       =  8  X  .  - 

180  2,880  04 


=  8  X  f  =  0  inches.     The  process,  however,  is  long-er  and  is 
not  so  direct,  and  the  first  method  is  to  be  preferred. 

155.  If  two  cylinders  have  equal  volumes,  but  different 
diameters,  the  diameters  are  to  each  other  inversely  as  the 
square  roots  of  their  lengths.  Hence,  if  it  is  desired  to  find 
the  diameter  of  a  cylinder  that  is  to  be  15  inches  long1,  and 
which  shall  have  the  same  volume  as  one  that  is  !)  inches 
in  diameter  and  12  inches  long-,  we  write  the  proportion 
0  :  x  =  VT5  :  4/T2. 

Since  neither  12  nor  15  are  perfect  squares,  we  square  all 
the  terms  (Arts.  154  and  143)  and  obtain 

81  :  .r2  =  15  :  12;  whence,  x*  =  —    —  =  (J4.8, 

Jy 

and     x  —    4/G4.8  =  8.05     inches  =  diameter     of     15-inch 
cylinder. 

EXAMPLES  FOR  PRACTICE. 

15G.      Solve  the  following  examples: 

1.  The  intensity  of  light  varies  inversely  as  the  square  of  the  dis- 
tance from  the  source  of  light.     If  a  gas  jet  illuminates  an  object  .50 
feet  away  with   a   certain  distinctness,  how  much   brighter   will   the 
object  be  at  a  distance  of  20  feet  ?  Ans.  2±  times  as  bright. 

2.  In  the  last  example,  suppose  that  the  object  had  been  40  feet 
from  the  gas  jet;  how  bright  would  it  have  been,  compared  with  its 
brightness  at  30  feet  from  the  gas  jet  ?  Ans.  /,,  as  bright. 

3.  When  comparing  one  light  with  another,  the  intensities  of  their 
illuminating  powers  vary  as  the  squares  of  their  distances  from  the 

1-8 


54  ARITHMETIC.  §  2 

source.  If  a  man  can  just  distinguish  the  time  indicated  by  his  watch, 
50  feet  from  a  certain  light,  at  what  distance  could  he  distinguish  the 
time  from  a  light  3  times  as  powerful  ?  Ans.  86.6+  feet. 

4.  The  quantity  of  air  flowing  through  a  mine  varies  directly  as  the 
square  root  of  the  pressure.     If  60,000  cubic  feet  of  air  flow  per  min- 
ute when  the  pressure  is  2.8  pounds  per  square  foot,  how  much  will 
flow  when  the  pressure  is  3.6  pounds  per  square  foot  ? 

Ans.  68,034  cu.  ft.  per  min.,  nearly. 

5.  In  the  last  example,  suppose  that  70,000  cubic  feet  per  minute 
had  been  required;    what  would  be  the  pressure  necessary  for  this 
quantity?  Ans.  3.81+  Ib.  per  sq.  ft. 


CAUSES  AND   EFFECTS. 

157.  Many  examples  in  proportion  may  be  more  easily 
solved  by  using  the  principle  of  cause  and  effect.     That 
which  may  be  regarded  as  producing  a  change  or  alteration 
in  something,  or  as  accomplishing  something,  may  be  called 
a  cause,  and  the  change  or  alteration,  or  thing  accomplished, 
as  the  effect. 

158.  Like  causes  produce  like  effects.     Hence,  when  two 
causes  of  the  same  kind  produce  two  effects  of  the  same 
kind,  the  ratio  of  the  causes  equals  the  ratio  of  the  effects ; 
in  other  words  the  first  cause  is  to  the  second  cause  as  the 
first  effect  is  to  the  second  effect.     Thus,  in  the  question — 
if  3  men  can  lift  1,400  pounds,  how  many  pounds  can  7 
men  lift? — we  call  3  men  and  7  men  the  causes  (since  they 
accomplish  something,  viz.,  the  lifting  of  the  weight),  the 
number  of  pounds  lifted,  viz.,  1,400  pounds  and  x  pounds, 
are  the  effects.    If  we  call  3  men  the  first  cause,  1,400  pounds 
is  the  first  effect ;  7  men  is  the  second  cause,  and  x  pounds 
is  the  second  effect.      Hence,  we  may  write 

1st  cause      2d  cause  1st  effect      2d  effect 

3:7        =       1,400       :       x 

7X1  400 

whence  x  —  ^ =  3,266f  pounds. 

o 

159.  The  principle  of  cause  and  effect  is   extremely 
useful  in  the  solution  of  examples  in  compound  proportion, 
as  we  shall  now  show. 


§  2  ARITHMETIC.  55 

COMPOUND  PROPORTION. 

16O.  All  the  cases  of  proportion  so  far  considered  have 
been  cases  of  simple  proportion  ;  i.  e. ,  each  term  has  been 
composed  of  but  one  number.  There  are  many  cases,  how- 
ever, in  which  two  or  all  the  terms  have  more  than  one 
number  in  them ;  all  such  cases  belong-  to  compound  pro- 
portion. In  all  examples  in  compound  proportion,  both 
causes  or  both  effects  or  all  four  consist  of  more  than  two 
numbers.  We  will  illustrate  this  by  an 

EXAMPLE. — If  40  men  earn  §1,280  in  16  days,  how  much  will  36  men 
earn  in  31  days? 

SOLUTION. — Since  40  men  earn  something,  40  men  is  a  cause,  and 
since  they  take  16  days  in  which  to  earn  something,  16  days  is  also  a 
cause.  For  the  same  reason  36  men  and  31  days  are  also  causes.  The 
effects,  that  which  is  earned,  are  1,280  dollars  and  x  dollars.  Then,  40 
men  and  16  days  make  up  the  first  cause,  and  36  men  and  31  days 
make  up  the  second  cause.  §1,280  is  the  first  effect,  and  §.r  is  the  second 
effect.  Hence,  we  write 

1st  cause  2d  cause          1st  effect   2d  effect 
40  36  ,  OHn 

16       '       31  1>38° 

Now,  instead  of  using  the  colon  to  express  the  ratio,  we  shall  use  the 
vertical  line  (see  Art.  118),  and  the  above  becomes 


40 


36 


16        31  = 

In  the  last  expression,  the  product  of  all  the  numbers  included 
between  the  vertical  lines  must  equal  the  product  of  all  the  numbers 
without  them  ;  i.  e.,  36  X  31  X  1,280  =  40  X  16  X  x  . 


161.  The  above  might  have  been  solved  by  canceling- 
factors  of  the  numbers  in  the  original  proportion.  For,  if 
any  number  within  the  lines  has  a  factor  common  to  any 
number  without  the  lines,  that  factor  may  be  canceled  from 
both  numbers.  Thus, 

2 

36  _    w 

31  ;^0 

16  is  contained  in  1,280,  80  times.     Cancel  1C  and  1,280,  and 
write  80  above  1,280.     40  is  contained  in  80,  2  times.     Cancel 


56 


ARITHMETIC. 


40  and  80,  and  write  2  above  80.  Now,  since  there  are  no  more 
numbers  that  can  be  canceled,  x  =  30x31x2  =  $2,232,  the 
same  result  as  was  obtained  in  the  preceding  article. 

16/2.  Rule. —  Write  all  t/ie  numbers  forming  the  first 
cause  in  a  vertical  column,' and  draiv  a  vertical  line ;  on  the 
other  side  of  this  line  write  in  a  vertical  column  all  the 
numbers  forming  the  second  cause.  Write  the  sign  of 
equality  to  the  right  of  the  second  column,  and  on  the  right 
of  this  form  a  third  column  of  the  numbers  composing  the 
first  effect,  drawing  a  vertical  line  to  the  right ;  on  the  other 
side  of  this  line,  ivrite  for  a  fourth  column,  the  numbers 
composing  the  second  effect.  There  must  be  as  many  num- 
bers in  the  second  cause  as  in  the  first  cause,  and  in  the 
second  effect  as  in  the  first  effect ;  hence,  if  any  term  is 
wanting,  write  x  in  its  place.  Multiply  togetlier  all  the 
numbers  within  the  vertical  lines,  and  also  all  those  without 
the  lines  (canceling  previously,  if  possible],  and  divide  the 
product  of  those  numbers  which  do  not  contain  x  by  the 
product  o"f  the  others  in  which  x  occurs,  and  the  result  tvill 
be  the  value  of  x. 

163.  EXAMPLE. — If  40  men  can  dig  a  ditch  720  feet  long,  5  feet 
wide,  and  4  feet  deep  in  a  certain  time,  how  long  a  ditch  6  feet  deep 
and  3  feet  wide  could  24  men  dig  in  the  same  time  ? 

SOLUTION. — Here  40  men  and  24  men  are  the  causes,  and  the  two 
ditches  are  the  effects.  Hence, 


24  = 


3  whence,  x  =  24  X  5  X  4  =  480  feet.     Ans. 


164.  EXAMPLE. — The  volume  of  a  cylinder  varies  directly  as  its 
length  and  directly  as  the  square  of  its  diameter.  If  the  volume  of  a  cylin- 
der 10  inches  in  diameter  and  20  inches  long  is  1,570.8  cubic  inches,  what  is 
the  volume  of  another  cylinder  16  inches  in  diameter  and  24  inches  long  ? 

SOLUTION. — In  this  example,  either  the  dimensions  or  the  volumes 
may  be  considered  the  causes;  say  we  take  the  dimensions  for  the 
causes.  Then,  squaring  the  diameters, 


102 
20 


=  1,570.8 


100       256 

24  =  1,570.8       x\ 
6 

whence,  x  =  — - — ? — '-  -  =  4,825.4976  cubic  inches.    Ans. 

o  X  J-00 


ARITHMETIC. 


165.  EXAMPLE. — If  a  block  of  granite  8  ft.  long,  5  ft.  wide,  and 
3  ft.  thick  weighs  7,200  lb.,  what  will  be  the  weight  of  a  block  of 
granite  12  ft.  long,  8  ft.  wide,  and  5  ft.  thick  ? 

SOLUTION. — Taking  the  weights  as  the  effects,  we  have 
4 


=  7,200 


.r,  or  .r  =  4  X  7,200  =  28,800  pounds,     Ans. 


166.  EXAMPLE. — If  12  compositors  in  BO  days  of  10  hours  eacli 
set  up  25  sheets  of  16  pages  each,  32  lines  to  the  page,  in  how  many 
days  8  -hours  long  can  18  compositors  set  up,  in  the  same  type,  04 
sheets  of  12  pages  each,  40  lines  to  the  page  ? 

SOLUTION. — Here  compositors,  days,  and  hours  compose  the  causes, 
and  sheets,  pages,  and  lines  the  effects.     Hence, 
33^2 


10 


VI,  or  x  =  ?>  X  10  X  2  =  C>0  days.     Ans. 


167.  In  examples  stated  like  that  in  Art.  164,  should 
an  inverse  proportion  occur,  write  the  various  numbers  as 
in  the  preceding  examples,  and  then  transpose  from  one 
side  of  the  vertical  line  to  the  other  side  those  numbers 
which  are  said  to  vary  inversely. 

EXAMPLE. — The  centrifugal  force  of  a  revolving  body  varies  directly 
as  its  weight,  as  the  square  of  its  velocity,  and  inversely  as  the  radius 
of  the  circle  described  by  the  center  of  the  body.  If  the  centrifugal 
force  of  a  body  weighing  15  pounds  is  187  pounds  when  the  body 
revolves  in  a  circle  having  a  radius  of  12  inches,  with  a  velocity  of  20 
feet  per  second,  what  will  be  the  centrifugal  force  of  the  same  body 
when  the  radius  is  increased  to  18  inches  and  the  speed  is  increased  to 
24  feet  per  second  ? 

SOLUTION. — Calling  the  centrifugal  force  the  effect,  we  have 


15 

20* 
12 


15 

24*  =  187 
18 


Transposing  12  and  18  (since  the  radii  are  to  vary  inversely)  and  squar- 
ing 20  and  24, 


=  187 


12 


,  or  x  = 


12X2X18 


-  =  179.52  pounds.     Ans. 


58  ARITHMETIC.  §  2 

EXAMPLES  FOR  PRACTICE. 

168.      Solve  the  following  by  compound  proportion: 

1.  If  12  men  dig  a  trench  40  rods  long  in  24  days  of  10  hours  each, 
how  many  rods  can  16  men  dig  in  18  days  of  9  hours  each  ? 

Ans.  36  rods. 

2.  If  a  piece  of  iron  7  feet  long,  4  inches  wide,  and  6  inches  thick 
weighs  600  pounds,  how  much  will  a  piece  of  iron  weigh  that  is  16  feet 
long,  8  inches  wide,  and  4  inches  thick  ?  Ans.  1,828|  Ib. 

3.  If  24  men  can  build  a  wall  72  rods  long,  6  feet  wide,  and  5  feet 
high  in  60  days  of  10  hours  each,  how  many  days  will  it  take  32  men 
to  build  a  wall  96  rods  long,  4  feet  wide,  and  8  feet  high,  working 
8  hours  a  day  ?  Ans.  80  days. 

4.  The  horsepower  of  an  engine  varies  as  the  mean  effective  pres- 
sure, as  the  piston  speed,  and  as  the  square  of  the  diameter  of  the 
cylinder.     If    an  engine    having  a  cylinder    14    inches  in  diameter 
develops  112  horsepower  when  the  mean  effective  pressure  is  48  pounds 
per  square  inch  and  the  piston  speed  is  500  feet  per  minute,  what  horse- 
power will  another  engine  develop .  if  the  cylinder  is   16  inches  in 
diameter,  piston  speed  is  600  feet  per  minute,  and  mean  effective  pres- 
sure is  56  pounds  per  square  inch  ?  Ans.  204.8  horsepower. 

5.  Referring  to  the  example  in  Art.  164,  what  will  be  the  volume 
of  a  cylinder  20  inches  in  diameter  and  24  inches  long  ? 

Ans.  7,539.84  cubic  inches. 

6.  Knowing  that  the  product  of  3x5x7x9  is  945,  what  is  the 
product  of  6  X  15  X  14  X  36  ?  Ans.  45,360. 


FORMULAS. 


1.  The  term  formula,  as  used  in  mathematics  and  in 
technical  books,  may  be  defined  as  a  rule  in   ivliicli  symbols 
are   used  instead  of  words;    in    fact,    a   formula   may    be 
regarded  as  a  shorthand  method  of  expressing  a  rule.     Any 
formula  can  be  expressed  in  words,  and  when  so  expressed 
it  becomes  a  rule. 

2.  Formulas    are    much    more   convenient   than    rules; 
they  show  at  a  glance  all  the  operations  that  are  to  be  per- 
formed ;  they  do  not  require  to  be  read  three  or  four  times, 
as  is  the  case  with  most  rules,  to  enable  one  to  imderstand 
their  meaning;   they  take  up  much  less  space,  both  in  the 
printed  book  and  in  one's  note  book,  than  rules;  in  short, 
whenever  a  rule  can  be  expressed  as  a  formula,  the  formula 
is  to  be  preferred. 

3.  As  the  term  "quantity"  is  a  very  convenient  one  to 
use,  we  will  define  it.     In  mathematics,  the  word  quantity 
is  applied  to  anything  that  it  is  desired  to  subject  to  the 
ordinary  operations  of  addition,  subtraction,  multiplication, 
etc.,  when  we  do  not  wish  to  be  more  specific  and   state 
exactly  what  the  thing  is.     Thus,  we  can  say  "two  or  more 
numbers,"  or  "two  or  more  quantities";  the  word  quantity 
is  more  general  in  its  meaning  than  the  word  number. 

4.  The  -signs  used  in  formulas  are  the  ordinary  signs 
indicative  of  operations,  and  the  signs  of  aggregation.      All 
these  signs  are  explained  in  arithmetic,  but  some  of  them 
will  here  be  explained  in  order  to  refresh  the  student's 
memory. 


2  FORMULAS.  §  3 

5.  The  signs  indicative  of  operations  are  six  in  number; 
viz.,  +  ,  — ,  X,  -5-,    |   ,  I/- 
Division is  indicated  by  the  sign  -^-,or  by  placing  a  straight 

25 
line  between  the  two  quantities.  Thus,  25  |  17,  25  /  17,  and-— 

all  indicate  that  25  is  to  be  divided  by  17.  When  both  quan- 
tities are  placed  on  the  same  horizontal  line,  the  straight 
line  indicates  that  the  quantity  on  the  left  is  to  be  divided 
by  that  on  the  right.  When  one  quantity  is  below  the 
other,  the  straight  line  between  indicates  that  the  quantity 
above  the  line  is  to  be  divided  by  the  one  below  it. 

The  sign  (|/ )  indicates  that  some  root  of  the  quantity  to 
the  right  is  to  be  taken ;  it  is  called  the  radical  sign.  To 
indicate  what  root  is  to  be  taken,  a  small  figure,  called  the 
index,  is  placed  within  the  sign,  this  being  always  omitted 
when  the  square  root  is  to  be  indicated.  Thus,  ^/25  indi- 
cates that  the  square  root  of  25  is  to  be  taken;  ^25  indicates 
that  the  cube  root  of  25  is  to  be  taken;  etc. 

6.  The  signs  of  aggregation  are  four  in  number  ;  viz. , 
()>    []»    I  l>   respectively  called   the   vinculum,   the 


parenthesis,  the  brackets,  and  the  brace  ;  they  are  used 
when  it  is  desired  to  indicate  that  all  the  quantities  included 
by  them  are  to  be  subjected  to  the  same  operation.  Thus, 
if  we  desire  to  indicate  that  the  sum  of  5  and  8  is  to  be 
multiplied  by  7,  and  we  do  not  wish  to  actually  add  5  and  8 
before  indicating  the  multiplication,  we  may  employ  any 


one  of  the  foiir  signs  of  aggregation  as  here  shown  :  5  +  8x7, 
(5  +  8)  X  7,  [5  +  8]  X  7,  {  5  +  8  }  X  7.  The  vinculum  is  placed 
above  those  quantities  which  are  to  be  treated  as  one 
quantity  and  subjected  to  the  same  operation. 

7.  While  any  one  of  the  four  signs  may  be  used  as 
shown  above,  custom  has  restricted  their  use  somewhat. 
The  vinculum  is  rarely  used  except  in  connection  with  the 
radical  sign.  Thus,  instead  of  writing  ^(5  +  8),  ^[5  +  8], 
or  ^{5  +  8}  for  the  cube  root  of  5  plus  8,  all  of  which  would 
be  correct,  the  vinculum  is  nearly  always  used,  ^5  -f-8. 

In   cases  where  but  one  sign  of  aggregation   is  needed 


§  3  FORMULAS.  3 

(except,  of  course,  when  a  root  is  to  he  indicated),  the  paren- 
thesis is  always  used.  Hence,  (5  -{-  8)  X  7  would  he  the 
usual  way  of  expressing  the  product  of  5  plus  8,  and  7. 

If  two  signs  of  aggregation  are  needed,  the  hrackets  and 
parenthesis  are  used,  so  as  to  avoid  having  a  parenthesis 
within  a  parenthesis,  the  brackets  being  placed  outside.  For 
example,  [(20  —  5)  -4-  3]  X  9  means  that  the  difference  between 
20  and  5  is  to  be  divided  by  3,  and  this  result  multiplied  by  '.». 

If  three  signs  of  aggregation  are  required,  the  brace, 
brackets,  and  parenthesis  are  used,  the  brace  being  placed 
outside,  the  brackets  next,  and  the  parenthesis  inside.  For 
example,  •  [(20  —  5)  -4-  3]  X  0  —  21 1  -4-8  means  that  the  quotient 
obtained  by  dividing  the  difference  between  20  and  r>  by  3 
is  to  be  multiplied  by  0,  and  that  after  21  has  been  subtracted 
from  the  product  thus  obtained,  the  result  is  to  be  divided  by  8. 

Should  it  be  necessary  to  use  all  four  of  the  signs  of  aggre- 
gation, the  brace  would  be  put  outside,  the  brackets  next, 
the  parenthesis  next,  and  the  vinculum  inside.  For 
example,  {[(20-5  -4-  3)  X  9  -  21]  -4-  8}  X  12. 

8.  As  stated  in  arithmetic,  when  several  quantities  are 
connected  by  the  various  signs  indicating  addition,  subtrac- 
tion, multiplication,  and    division,    the    operation    indicated 
by  the  sign  of  multiplication  must  always  be  performed  first. 
Thus,  2  +  3  X  4  is  equal  to  14,  3  being  multiplied  by  4,  before 
adding  to  2.     Similarly,  10-^2x5  is  equal  to  1,  since  2x~> 
equals  10,  and  10 -=-10  is  equal  to  1.      Hence,  in  the  above 
case,  if   the   brace   were    omitted,  the   result   would    be    \, 
whereas,  by  inserting  the  brace,  the  result  is  30. 

Following  the  sign  of  multiplication  comes  the  sign  of 
division  in  order  of  importance.  For  example,  5  —0-4-3  is 
equal  to  2,  9  being  divided  by  3  before  subtracting  from  ~>. 
The  signs  of  addition  and  subtraction  arc  of  equal  value  ; 
that  is,  if  several  quantities  are  connected  by  plus  and  minus 
signs,  the  indicated  operations  may  be  performed  in  the 
order  in  which  the  quantities  are  placed. 

9.  There  is  one  other  sign  used,  which  is  neither  a  sign 
of  aggregation  nor  a  sign  indicative  of  an  operation  to  be 


4  FORMULAS.  §  3 

performed  ;  it  is  (  =  ),  and  is  called  the  sign  of  equality ;  it 
means  that  all  on  one  side  of  it  is  exactly  equal  to  all  on 
the  other  side.  For  example,  2  =  2,  5  —  3  =  2,  5  X  (14  —  9) 

=  25. 

1O.  Having  called  particular  attention  to  certain  signs 
used  in  formulas,  the  formulas  themselves  will  now  be 
explained.  First,  consider  the  well  known  rule  for  finding  the 
horsepower  of  a  steam  engine,  which  may  be  stated  as  follows : 

Divide  the  continued  product  of  the  mean  effective  pressure 
in  pounds  per  square  inch,  the  length  of  the  stroke  in  feet,  the 
area  of  the  piston  in  square  inches,  and  the  number  of  strokes 
per  minute,  by  33,000;  the  result  will  be  the  horsepower. 

This  is  a  very  simple  rule,  and  very  little,  if  anything,  will 
be  saved  by  expressing  it  as  a  formula,  so  far  as  clearness  is 
concerned.  The  formula,  however,  will  occupy  a  great  deal 
less  space,  as  we  shall  show. 

An  examination  of  the  rule  will  show  that  four  quantities 
(viz.,  the  mean  effective  pressure,  the  length  of  the  stroke, 
the  area  of  the  piston,  and  the  number  of  strokes)  are  multi- 
plied together,  and  the  result  is  divided  by  33,000.  Hence, 
the  rule  might  be  expressed  as  follows : 

TT  mean  effective  pressure     .,  stroke 

~  (in  pounds  per  square  inch)  x  (in  feet) 

..     area  of  piston      .,  number  of  strokes  _._  go  QQQ 
*  (in  square  inches)  x        (per  minute) 

This  expression  could  be  shortened  by  representing  each 
quantity  by  a  single  letter;  thus,  representing  horsepower 
by  the  letter  "//,"  the  mean  effective  pressure  in  pounds 
per  square  inch  by  '"/*,"  the  length  of  stroke  in  feet  by  "Z, " 
the  area  of  the  piston  in  square  inches  by  "A,"  the  number 
of  strokes  per  minute  by  '  W, "  and  substituting  these  letters 
for  the  quantities  that  they  represent,  the  above  expression 
would  reduce  to 

H  = 


33,000 

a  much  simpler  and  shorter  expression.     The  last  expression 
is  called  a  formula. 


§  3  FORMULAS.  5 

11.  The  formula  just  given  shows,  as  we  stated  in  the 
beginning,  that  a  formula  is  really  a  shorthand  method  of 
expressing  a  rule.  It  is  customary,  however,  to  omit  the 
sign  of  multiplication  between  two  or  more  quantities  when 
they  are  to  be  multiplied  together,  or  between  a  number  and 
a  letter  representing  a  quantity,  it  being  always  understood 
that,  when  two  letters  are  adjacent,  with  no  sign  between 
them,  the  quantities  represented  by  these  letters  are  to  be 
multiplied.  Bearing  this  fact  in  mind,  the  formula  just 
given  can  be  further  simplified  to 

PLAN 


H  = 


33,000 


The  sign  of  multiplication,  evidently,  cannot  be  omitted 
between  two  or  more  numbers,  as  it  would  then  be  impos- 
sible to  distinguish  the  numbers.  A  near  approach  to  this, 
however,  may  be  attained  by  placing  a  dot  between  the 
numbers  which  are  to  be  multiplied  together,  and  this  is  fre- 
quently done  in  works  on  mathematics  when  it  is  desired 
to  economize  space.  In  such  cases  it  is  usual  to  put  the  dot 
higher  than  the  position  occupied  by  the  decimal  point. 
Thus  2-3  means  the  same  as  2x3;  542-749 -1,OOG  indicates 
that  the  numbers  542,  749,  and  1,OOG  are  to  be  multiplied 
together. 

It  is  also  customary  to  omit  the  sign  of  multiplication  in 
expressions  similar  to  the  following:  a  X  Vb  -\-  c,  3x(/'  +  0> 
(b  +  c]  X  a,  etc. ,  writing  them  aVb  -f-  c,  3  (b  -f-  c],  (b  +  c]  a,  etc. 
The  sign  is  not  omitted  when  several  quantities  are  included 
by  a  vinculum,  and  it  is  desired  to  indicate  that  the  quanti- 
ties so  included  are  to  be  multiplied  by  another  quantity. 
For  example,  3  X  b  +  c,  b  +  c  X  a,  Vb  +  cXa,  etc.  are  always 
written  as  here  printed. 

12.  Before  proceeding  further,  we  will  explain  one  other 
device  that  is  .used  by  formula  makers,  and  which  is  likely  to 
puzzle  one  who  encounters  it  for  the  first  time — it  is  the  use  of 
what  mathematicians  call  primes  and  subs. ,  and  what  printers 
call  superior  and  inferior  characters.  As  a  rule,  formula 
makers  designate  quantities  by  the  initial  letters  of  the  names 


6  FORMULAS.  §  3 

of  the  quantities.  For  example,  they  represent  volume  by  v, 
pressure  by  /,  height  by  //,  etc.  This  practice  is  to  be  com- 
mended, as  the  letter  itself  serves  in  many  cases  to  identify 
the  quantity  which  it  represents.  Some  authors  carry  the 
practice  a  little  further,  and  represent  all  quantities  of  the 
same  nature  by  the  same  letter  throughout  the  book,  always 
having  the  same  letter  represent  the  same  thing.  Now,  this 
practice  necessitates  the  use  of  the  primes  and  subs,  above 
mentioned,  when  two  quantities  have  the  same  name  but 
represent  different  things.  Thus,  consider  the  word  pressure 
as  applied  to  steam,  at  different  stages  between  the  boiler 
and  the  condenser.  First,  there  is  absolute  pressure,  which 
is  equal  to  the  gauge  pressure  in  pounds  per  square  inch  plus 
the  pressure  indicated  by  the  barometer  reading  (usually 
assumed  in  practice  to  be  14.7  pounds  per  square  inch,  when 
a  barometer  is  not  at  hand).  If  this  be  represented  by  /, 
how  shall  we  represent  the  gauge  pressure  ?  Since  the  abso- 
lute pressure  is  always  greater  than  the  gauge  pressure,  sup- 
pose we  decide  to  represent  it  by  a  capital  letter,  and  the 
gauge  pressure  by  a  small  (lower-case)  letter.  Doing  so,  P 
represents  absolute  pressure,  and  /,  gauge  pressure.  Fur- 
ther, there  is  usually  a  "drop"  in  pressure  between  the 
boiler  and  the  engine,  so  that  the  initial  pressure,  or  pressure 
at  the  beginning  of  the  stroke,  is  less  than  the  pressure  at 
the  boiler.  How  shall  we  represent  the  initial  pressure  ? 
We  may  do  this  in  one  of  three  ways  and  still  retain  the 
letter  p  or  P  to  represent  the  word  pressure :  First,  by  the 
use  of  the  prime  mark ;  thus,  p'  or  P'  (read  /  prime  and 
P  major  prime)  may  be  considered  to  represent  the  initial 
gauge  pressure,  or  the  initial  absolute  pressure.  Second,  by 
the  use  of  sub.  figures ;  thus,  /t  or  Pl  (read  /  sub.  one,  and 
P  major  sub.  one}.  Third,  by  the  use  of  sub.  letters;  thus, 
/,  or  P{  (read  p  sub.  i  and  P  major  sub.  i).  In  the  same 
manner/"  (read/  second),  /2,  or/r  might  be  used  to  repre- 
sent the  gauge  pressure  at  release,  etc.  The  sub.  letters 
have  the  advantage  of  still  further  identifying  the  quantity 
represented ;  in  many  instances,  however,  it  is  not  convenient 
to  use  them,  in  which  case  primes  and  subs,  are  used  instead. 


§  3  FORMULAS.  7 

The  prime  notation  may  be  continued  as  follows:  />'",  p*\  p\ 
etc. ;  it  is  inadvisable  to  use  superior  figures,  for  exam- 
ple,/\  />2,  />3,  /",  etc.,  as  they  are  liable  to  be  mistaken  for 
exponents. 

13.  The  main  thing  to  be  remembered  by  the  student  is 
that  iv hen  a  formula  is  green  in  ivhieh  the  same  letters  oeenr 
several  times,  all  like  letters  /taring  the  same  primes  or  subs, 
represent  the  same  quantities,  icliile  those  ivhieh  differ  in  any 
respect  represent  different  quantities.  Thus,  in  the  formula 


wv  iv.v  and  ti'3  represent  the  weights  of  three  different 
bodies;  sv  s.2,  and  jr3,  their  specific  heats;  and  tv  /.„  and  /.., 
their  temperatures;  while  /  represents  the  final  temperature 
after  the  bodies  have  been  mixed  together.  It  should  be 
noted  that  those  letters  having  the  same  subs,  refer  to  the 
same  bodies.  Thus,  u'r  sv  and  t^  all  refer  to  one  of  the  three 
bodies;  ivr  sr  /2,  to  another  body;  etc. 

14.  It  is  very  easy  to  apply  the  above  formula  when  the 
values  of  the  quantities  represented  by  the  different  letters 
are  known.  All  that  is  required  is  to  substitute  the  numer- 
ical values  of  the  letters,  and  then  perform  the  indicated 
operations.  Thus,  suppose  that  the  values  of  ^\,  sv  tl  are, 
respectively,  2  pounds,  .0051,  and  80°;  of  ?i',,  s2,  and  /,,  7.8 
pounds,  1,  and  80°;  and  of  tt'3,  ^3,  and  /3,  3|  pounds,  .1138, 
and  780°;  then,  the  final  temperature  /  is,  substituting  these 
values  for  their  respective  letters  in  the  formula, 

_  2  X  .0051  X  80  +  7.8  X  1  X  80  +  3|  X  .1 138  X  780 

2X.0051  +  7.8Xl  +  3iX.1138 
_  15.216  +  624  +  288.483  __  027.600  _  0 

.1902 +  7. 8 +  .30085       ~  8.30005  ~ 

In  substituting  the  numerical  values,  the  signs  of  multi- 
plication are,  of  course,  written  in  their  proper  places;  all 
the  multiplications  are  performed  before  adding,  according 
to  the  rule  previously  given. 


8  FORMULAS.  §  3 

15.  The  student  should  now  be  able  to  apply  any  formula 
involving  only  algebraic  expressions  that  he  may  meet  with, 
and  which  do  not  require  the  use  of  logarithms  for  their 
solution.  We  will,  however,  call  his  attention  to  one  or  two 
other  facts  that  he  may  have  forgotten. 

Expressions  similar  to  — —  sometimes  occur,  the  heavy  line 
ooO 

~25~ 

indicating  that  160  is  to  be  divided  by  the  quotient  obtained 
by  dividing  660  by  25.     If  both  lines  were  light  it  would  be 

/>  />  r\ 

impossible  to  tell  whether  160  was  to  be  divided  by  — — ,  or 

<i& 
~\  f*f\ 

whether  — —  was  to  be  divided  by  25.     If  this  latter  result 
660 

160 


were   desired,  the   expression   would   be   written   —  —  .     In 

lilU 

every  case,  the  heavy  line  indicates  that  all  above  it  is  to  be 

divided  by  all  below  it. 

i  PO 
In  an  expression  like  the  following,  -  the  heavy 


line  is  not  necessary,  since  it  is  impossible  to  mistake  the 
operation   that   is   required   to   be    performed.      But,  since 


660        175  +  660    .,  175  +  660,          .  660 

w  :    nfi—  >  lf  we  substltute  -  -25-  -  f  or  7  +  w 

the   heavy  line   becomes   necessary  in  order  to  make  the 
resulting  expression  clear.     Thus, 

160  160  160^ 

660  ~"  175  +  660  ~  835~' 
~25~  25  "25" 

16.  Fractional  exponents  are  sometimes  used  instead  of 
the  radical  sign.  That  is,  instead  of  indicating  the  square, 
cube,  fourth  root,  etc.  of  some  quantity,  as  37  by  4/37, 
f37,  f37,  etc.,  these  roots  are  indicated  by  37*,  37*  37*, 
etc.  Should  the  numerator  of  the  fractional  exponent  be 
some  quantity  other  than  1,  this  quantity,  whatever  it  may 


§  3  FORMULAS.  9 

be,  indicates  that  the  quantity  affected  by  the  exponent  is 
to  be  raised  to  the  power  indicated  by  the  numerator;  the 
denominator  is  always  the  index  of  the  root.  Hence, 
instead  of  writing  &3T2  for  the  cube  root  of  the  square  of  37, 
it  may  be  written  37^,  the  denominator  being  the  index  of 
the  root;  in  other  words,  ^372  =  37s.  Likewise,  V(l+rtY>):< 
may  also  be  written  (1  +  (ilb)*,  a  much  simpler  expression. 

17.  We  will  now  give  several  examples  showing  how  to 
apply  some  of  the  more  difficult  formulas  that  the  student 
may  encounter. 

The  area  of  any  segment  of  a  circle  that  is  less  than  (or 
equal  to)  a  semicircle  is  expressed  by  the  formula 

rl  E      c  , 


^== 


865 


in  which  A  =  area  of  segment  ; 

TT    =  3.1416; 

r    =  radius; 

E  =  angle    obtained   by   drawing  lines  from   the 
center  to  the  extremities  of  arc  of  segment  ; 

c    —  chord  of  segment; 
and  h    =  height  of  segment. 

EXAMPLE.—  What  is  the  area  of  a  segment  whose  chord  is  10  inches 
long,  angle  subtended  by  chord  is  83.46°,  radius  is  7.5  inches,  and 
height  of  segment  is  1.91  inches  ? 

SOLUTION.  —  Applying  the  formula  just  given, 

.        7rr*£      c.  3.  1416  X  7.  5*  X  83.  46      10 

^=-360—  2^~^=  -1360-  -T(7.5-1.91) 

=  40.968  —  27.95  =  13.018  square  inches,  nearly.     Ans. 

18.  The  area  of  any  triangle  may  be  found  by  means  of 
the  following  formula,  in  which  A  =  the  area,  and  a,  b,  and 
c  represent  the  lengths  of  the  sides  : 


EXAMPLE. — What  is  the  area  of  a  triangle  whose  sides  are  21  feet. 
46  feet,  and  50  feet  long  ? 


10  FORMULAS.  §  3 

SOLUTION.  —  In  order  to  apply  the  formula,  suppose  we  let  a  repre- 
sent the  side  that  is  21  feet  long;  b,  the  side  that  is  50  feet  long;  and 
c  ,  the  side  that  is  46  feet  long.  Then,  substituting  in  the  formula, 


b    f~      /tf'  +  ^-^1      50 
A  = 


2X50 


=  25|/441 -8. 25*  =  25 4/44 1  -  68.0625  =  25|/372.9375 
—  25  X  19.312  =  482.8  square  feet,  nearly.     Ans. 

19.  The  operations  in  the  above  examples  have  been 
extended  much  farther  than  was  necessary ;  it  was  done  in 
order  to  show  the  student  every  step  of  the  process.     The 
last   formula   is   perfectly  general,   and    the    same    answer 
would  have  been  obtained  had  the  50-foot  side  been  repre- 
sented by  a,  the  46-foot  side  by  b,  and  the  21-foot  side  by  c. 

20.  The    Rankine  -  Gordon    formula    for    determining 
the  least  load  in  pounds  that  will  cause  a  long  column  to 

break  is      " 

p__SA_ 

J-       — —  , .,  ' 


in  which  P  =  load  (pressure)  in  pounds;' 

vS"  =  iiltimate  strength  (in  pounds  per  square  inch) 

of  the  material  composing  the  column ; 
A  =  area   of    cross-section    of   column   in    square 

inches ; 

q  =  a   factor    (multiplier)   whose   value    depends 
upon  the  shape  of  the  ends  of  the  column  and 
on  the  material  composing  the  column; 
/  =  length  of  column  in  inches ; 

and          G  =  least  radius  of  gyration   of  cross-section  of 
column. 

The  values  of  S,  g,  and  Gl  are  given  in  printed  tables  in 
books  in  which  this  formula  occurs. 

EXAMPLE. — What  is  the  least  load  that  will  break  a  hollow  wrought- 
iron  column  whose  outside  diameter  is  14  inches;  inside  diameter, 
11  inches;  length,  20  feet;  and  whose  ends  are  flat  ? 


§3 


FORMULAS. 


11 


SOLUTION.—  For  steel,  5  =  150,000,  and  for  flat-ended  steel  col- 
umns, q  —  .-,-  U0():  A,  the  area  of  the  cross-section,  =  .  7854  (</,-'  —  ,/.,-'  ) 
—  .7854(142  —  11-),  </!  and  <f.,  being  the  outside  and  inside  diameters, 
respectively;  /  =  20  X  12  =  240  inches;  and  u-  =  '/l"h//-'  =  14'  +.  n'. 
Substituting  these  values  in  the  formula, 

p  _       SA        _  150.000  x  .7854(14-  -  11-) 

^T7"  ~~  ~T~     ~~2lu--i 

+  'r--- 


150,000  X  58.905        8,835,750 


1  +  .1163 


1.11G3 


Itj 

pounds.     Ans. 


21.      EXAMPLE.—  When  A  =  10, 
is  the  value  of  £"  in  the  following  : 


=  8,  C  =  5,  and  D  =  4,  what 


zr 

E= 


SOLUTION. — (a)  Substituting, 
E  = 


To  simplify  the  denominator,  square  the  4  and  5,  add  the  resulting 
fraction  to  2,  and  multiply  by  10.     Simplifying,  we  have, 


E  = 


160 


10xi        ir 


.1  /!7>0  _     3 /•><><) 
iiiio  ~   \   "33" 


Reducing  the  fraction  to  a  decimal,  so  that  it  will  be  easier  to  extract 
the  cube  root, 


E  -  V  (UJ606  =  1.823.     Ans. 


Substituting, 


10  +  22 

=  7  +  17.066+  =   24.066+  =  3Q08+      ^ 
10  -  4/4  8 


1-9 


12  FORMULAS.  §  3 

EXAMPLES  FOR  PRACTICE. 

Find  the  numerical  values  of  x  in  the  following  formulas,  when 
A  =  9,  B  =  8,  d  =  10,  e  =  3,  and  c  =  2: 


,              a  +  ^ 

Ans. 
Ans. 
Ans. 
Ans. 
Ans.  x 
Ans.  x  — 

=  * 

c  e 
Ae          5 

x  =  29. 
x  =  2. 
—  12  A. 

:    .396+. 

/      Bed 
ft      v  —    ,  / 

GEOMETRY  AND  MENSURATION. 


GEOMETRY. 


AND   ANGTJES. 

1.  Geometry  is  that  branch  of  mathematics  which  treats 
of  the  properties  of  lines,  angles,  surfaces,  and  volumes. 

2.  A  point   indicates    position    only.       It    has   neither 
length,  breadth,  nor  thickness. 

3.  A  line  has  only  one  dimension :  length. 

4.  A    straight  line  is  one   that    does   not    change   its 
direction  throughout  its  whole  length.    See 

Fig.  1.     A  straight  line  is  also  frequently 
called  a  right  line. 

5.  A  curved  line  changes  its  direc- 
tion at  every  point.      See  Fig.  2. 

6.  A  "broken  line    is   one    made  up 
wholly  of  straight  lines  lying  in  different 
directions.      See  Fig.  3. 

7.  Parallel  lines  are  equally  distant 
from  each  other  at  all  points.      The  lines 
shown  in  Fig.  4  are  parallel. 

8.  A    line    is    perpendicular    to 
another  when  it  meets  that  line  so  as 
not  to  incline  towards  it  on  either  side. 
Thus,  in  Fig.  5,  the  line  denoted  by  the 
letters  A,  B  is  perpendicular   to    that   C- 
denoted  by  C,  D. 


Fin.  J. 


I) 


GEOMETRY  AND  MENSURATION. 


Horizontal. 

FIG.  6. 


9.  A  horizontal  line  is  a  line  parallel 
to  the  horizon,  or  water  level.     See  Fig.  6. 

10.  A  vei'tical  line  is  a  line  perpen- 
dicular to  a  horizontal  line ;  consequently, 
it  has  the  direction  of  a  plumb-line.     See 
Fig.  6. 


11.  When  two  lines  cross  or  cut 
each  other,  they  are  said  to  intersect, 
and  the  point  at  which  they  intersect,  as 
A,  Fig.  7,  is  called  the  point  of  inter- 
section. 


FIG.  7. 


12.  An  angle  is  the  opening  between 
two  lines  which  intersect,  or  meet;  the 
point  of  meeting  is  called  the  vertex  of 

the  angle.      See  Fig.  8. 


13.  In  order  to  distinguish  one  line  from  another,  two 
of  its  points  are  given  if  it  is  a  straight  line,  and  as  many 
more  as  are  considered  necessary  if  it  is  a  broken  or  curved 
line.  Thus,  in  Fig.  9,  the  line  A  B  would  mean  the  straight 
line  included  between  the  points  A  and  A 
B.  Similarly,  the  straight  line  between 
C  and  B,  or  between  B  and  D,  would 

be  called  the  line  C  B,  or  the  line  B  D.    c—  — D 

The  broken  line  made  up  of  the  lines  FIG.  9. 

A  B  and  C  B,  or  B  D,  would  be  called  the  broken  line  C  B  A 
or  A  B  C,  and  A  B  D  or  D  B  A,  according  to  the  point  started 
from. 

To  distinguish  angles,  a  point  on  each  line  and  the  point 
of  their  intersection,  or  vertex  of  the  angle,  are  named; 
thus,  in  Fig.  9,  the  angle  formed  by  the  lines  A  B  and  C  B 
is  called  the  angle  A  B  C  or  the  angle  C  B  A  ;  the  letter  at 
the  vertex  is  always  placed  in  the  middle.  The  angle  formed 
by  the  lines  A  B  and  B  D  is  called  the  angle  A  B  D  or  the 
angle  DBA. 

When  an  angle  stands  alone  so  that  it  cannot  be  mistaken 


GEOMETRY  AND  MENSURATION. 


for  any  other  angle,  only  the  vertex  letter  need  be  given; 
thus,  the  angle  O,  or  the  angle  J-\  etc. 

14.  If  one  straight  line  meets  another  straight  line  at  a 
point  between  its  ends,  two  angles,  A  B  6"  and  A  B  D,  Fig.  'j, 
are  formed,  which  are  called  adjacent  angles. 


15.  When  these  adjacent  angles, 
ABC  and  A  B  D,  are  equal,  they  are 
called  right  angles.  See  Fig.  10. 


-23 


1(5.  An  acute  angle  is  less  than  a 
right  angle.  Thus,  A  J>  C\  Fig.  11,  is 
an  acute  ansfle. 


FIG.  11. 


17.     An    obtuse   angle    is   greater  A 
than  a  right  angle.      The  angle  A  B  D, 
Fig.  12,  is  an  obtuse  angle. 


18.  When  two  straight  lines  intersect,  they  form  four 
angles  about  the  point  of  intersection.  Thus,  in  Fig.  13, 
the  lines  A  B  and  C  D,  intersecting  at  the  point  O,  form 
four  angles,  B  O  D,  DO  A, .A  O  C,  and 
COB,  about  the  point  O.  The  angles 
which  lie  on  the  same  side  of  one 
straight  line,  as  DOB  and  D  O  A  are 
adjacent  angles.  The  angles  which 
lie  opposite  each  other  are  called  oppo- 
site angles.  Thus,  A  O  C  and  DOB,  also  DO  A  and 
B  O  C,  are  opposite  angles. 

When  one  straight  line  intersects  another  straight  line,  as 
in  Fig.  13,  the  opposite  angles  are  equal.  Thus,  D  O  B 
-  A  O  C,  and  D  OA  =  B  O  C. 


FIG.  13. 


GEOMETRY  AND  MENSURATION. 


B 

FIG.  14. 


19.     When  one  straight  line  meets 

another  straight  line  at  a  point  between 

its  ends,  the  sum  of  the  two  adjacent 

-•*>  angles,   as  A  B  D  and  ABC,  Fig.  14, 

is  equal  to  two  right  angles. 


2O.  If  a  number  of  straight  lines 
on  the  same  side  of  a  given  straight 
line  meet  at  the  same  point,  "the  sum  of 
all  the  angles  formed  is  equal  to  two 
right  angles.  Thus,  in  Fig.  15,  COB 
+  DO  C+E  O  D  +  FO  E  +  'A  OF 
=  two  right  angles. 


O 

FIG.  15. 


D 

FIG.  16. 


If  a  straight  line  intersects  another  straight  line,  so 
c  that  the  adjacent  angles  are  equal,  the 

lines  are  said  to  be  perpendicular  to 
each  other.  In  such  a  case,  four  right 
angles  are  formed  about  the  point  of 
intersection.  Thus,  in  Fig.  16,  B  O  C 
=  C  O  A;  hence,  B  O  C,  C  O  A,  A  O  D, 
and  DOB  are  right  angles.  From 
this,  it  is  seen  that  four  right  angles 
are  all  that  can  be  formed  about  a  given  point. 

It  follows  that,  if  through  a  given 
point,  any  number  of  straight  lines  are 
drawn,  the  sum  of  all  the  angles 
formed  about  the  point  of  intersection 
is  equal  to  four  right  angles.  Thus,  in 
Fig.  17,  H  O  F+  F  O  C+  C  O  A  +  A  OG 
+  GOE  +  EOD  +  DOB  +  BOH 
=  four  right  angles. 

EXAMPLE. — A  circular  window  has  12  ribs  equally  spaced.  What 
part  of  a  right  angle  is  included  between  the  center  lines  of  any  two 
ribs? 

SOLUTION. — Since  there  are  12  ribs,  there  are  12  angles.  The  sum  of 
all  the  angles  equals  four  right  angles.  Hence,  one  angle  equals  T\  of 
four  right  angles,  or  T4S  =  1  of  one  right  angle.  Ans. 


GEOMETRY  AND  MENSURATION. 


22.  A  perpendicular  drawn  from  a 
point  over  or  under  a  given  straight 
line  is  the  shortest  distance   from  the 
point    to    the    line,    or     to     the     line 
extended.     Thus,   if  A,  Fig.  18,  is  the 
given  point,   and   CD,  the  given  line, 
then    the    perpendicular     A  B    is    the 
shortest  distance  from  A  to  CD. 

23.  If  two  angles  have  their  sides 
parallel,  and  lie  in  the  same  or  in  oppo- 
site directions,  they  are  equal.     Thus, 
if  the    side   A  B,  Fig.  1!)  or  Fig.  20,  is 
parallel  to  the  side  D  E,  and  if  the  side 
B  C  is  parallel  to  the  side  E  F,  then  the 
angle  E  =  the  angle  B. 


Pic..  20. 


24.  If  two  sides  of  an 
angle  are  perpendicular  to 
two  sides  of  another  angle, 
the  two  angles  are  equal. 
Thus,  if  DE  and  G  If, 
Fig.  21,  are  perpendicular 
to  B  A,  and  E  F  and  //  A' 
are  perpendicular  to  B  C, 
then  will  angle  E  =  angle 
B  =  angle  //. 


EXAMPLES  YOU  PRACTICE. 

25.     Solve  the  following  examples: 

1.  In  a  pulley  with  five  arms,  what  part  of  a  right  angle  is  included 
between  the  center  lines  of  any  two  arms  ?          Ans.  i  of  a  right  angle. 

2.  If  one  straight  line  meets  another  straight  line  so  as  to  form  an 
angle  equal  to  If  right  angles,  what  part  of  a  right  angle  docs  its  adja- 
cent angle  equal  ?  Ans.  f  of  a  right  angle. 


6  GEOMETRY  AND  MENSURATION.  §  4 

3.  If  a  number  of  straight  lines  meet  a  given  straight  line  at  a 
given  point,  all  being  on  the  same  side  of  the  given  line,  so  as  to  form 
six  equal  angles,  what  part  of  a  right  angle  is  contained  in  each  angle  ? 

Ans.  £  of  a  right  angle. 


PLANE    FIGURES. 

26.  A   surface   has  only  two   dimensions:  length   and 
breadth. 

27.  A  plane  surface  is  a  flat  surface.      If  a  straightedge 
be  laid  on  a  plane  surface,  every  point  along  the  edge  of  the 
straightedge   will   touch   the   surface,    no  matter  in   what 
direction  it  is  laid. 

28.  A   plane   figure  is  any  part   of   a   plane   surface 
bounded  by  straight  or  curved  lines. 

29.  When  a  plane  figure  is  bounded  by  straight  lines,  it 
is  called  a  polygon.     The  bounding  lines  are  called  the  sides, 
and  the  length  of  the  broken  line  that  bounds  it  (or  the  whole 
distance  around  it)  is  called  the  perimeter  of  the  polygon. 

3O.  The  angles  formed  by  the  sides 
are  called  the  angles  of  the  polygon. 
Thus,  A  B  CD  E,  Fig.  22,  is  a  polygon. 
A  B,  B  C,  etc.  are  the  sides ;  E  A  B, 
BCD,  etc.  are  the  angles;  and  the 
length  of  the  broken  line  A  B  C  D  E  is  the 
perimeter. 

31.  Polygons  are  classified  according  to  the  number  of 
their  sides :  One  of  three  sides  is  called  a  triangle  ;  one  of 
four  sides,  a  quadrilateral ;  one  of  five  sides,  a  pentagon  ; 
one  of  six  sides,  a  hexagon ;    one  of  seven  sides,  a  hep- 
tagon ;  one  of  eight  sides,  an  octagon ;  one  of  ten  sides, 
a  decagon  ;  one  of  twelve  sides,  a  dodecagon  ;  etc. 

32.  Equilateral   polygons  are 

those  in  which  the  sides  are  all  equal. 
Thus,  in  Fig.  23,  A  B  =  B  C  =  CD 
=  DA  ;  hence,  A  BCD  is  an  equi- 
lateral  polygon.  FIG.  23. 


§  4  GEOMETRY  AND  MENSURATION. 

33.      An   equiangular  polygon   is  -A 

one  in  which  all  the  angles  are  equal. 
Thus,  in  Fig.  24,  angle  A  =  angle  B 
—  angle  D  =  angle  C  ;  hence,  A  B  D  C 
is  an  equiangular  polygon. 

34.     A  regular  polygon   is  one   in 

which  all  the  sides  and  all  the  angles  are 
equal.  Thus,  in  Fig.  25,  ;l  B  =  B  D 
=  DC—  C  A,  and  angle  A  —  angle  B 
=  angle  D  =  angle  C;  hence,  .-I  B  D  C 
is  a  regular  polygon. 
Other  regular  polygons  are  shown  in  Fig.  20. 


Pentagon.        Hexagon.        Heptagon.         Octagon.  Decagon.        Dodecagon. 

FIG.  2<i. 

35.     The  sum  of  all  the  interior  angles  of  any  polygon 
is  equal  to  two  right  angles  multiplied 
by  a  number  which  is  two  less  than  the 
number  of  sides  in  the  polygon.      Thus, 
A  B  CD  E  F,  Fig.  27,  is  a  polygon  of  six  F 
sides  (hexagon),  and  the  sum  of  all  the 
interior   angles  A+K+C+D  +  R  +  F 
=  2    right    angles  X  4    (=  G  — 2),    or    8 
right  angles. 

EXAMPLE. — Fig.  27  represents  a  regular  hexagon  (lias  equal  sides  and 
equal  angles).  How  many  right  angles  are  there  in  each  interior  angle? 

SOLUTION. — The  sum  of  the  interior  angles  is  2  X  (<>  —  ~)  =  N  right 
angles;  and,  as  there  are  six  equal  angles,  we  have  H  -=-  0  =  I.1,  right 
angles,  the  number  of  right  angles  in  each  interior  angle.  Ans. 


THE    TRIANGLE. 

36.  Triangles  may  be  divided,  with  respect  to  their  sides, 
into  isosceles,  equilateral,  and  scalene  triangles;  and  with 
respect  to  their  angles,  into  right-angled  and  oblique-angled 
triangles. 


GEOMETRY  AND  MENSURATION. 


37.     An  Isosceles  triangle  is  one  having  two 
of  its  sides  equal.     See  Fig.  28. 


FIG.  as. 


38.     An  equilateral  triangle  is  one  that 
has  the  three  sides  equal.      See  Fig.  29. 


FIG.  29. 


FIG.  so. 


39.     A  scalene  triangle  is  one  having  no 
two  of  its  sides  equal.      See  Fig.  30. 


4O.  A  right-angled  triangle  is  any  triangle  having 
one  right  angle.  See  Fig.  31.  The  side 
opposite  the  right  angle  is  called  the  hypot- 
enuse. For  brevity,  a  right-angled  tri- 
angle is  usually  termed  a  right  tri- 
angle. 


FIG.  31. 


41.     An  oblique-angled  or  oblique  triangle  is 
one  which  has  no  right  angles.     See  Fig.  32. 


FIG.  32. 

42.     The  base  of  any  triangle  is  the  side  upon  which  the 
triangle  is  supposed  to  stand. 

The  altitude  of  any  triangle  is  a  line  drawn  from  the  vertex 
of  the  angle  opposite  the 
base  perpendicular  to  the 
base   or  to   the  base  ex- 
tended.   Thus,  in  Figs.  33 
and  34,  the  side  A  C  is  the 
O  base  of  the  triangle  and 
FIG.  83.  the  line  BD  is  the  altitude. 

In  a  right  triangle,  if  one  of  the  short  sides  is  taken  as 


§4  GEOMETRY  AND   MENSURATION.  ji 

the  base,  the  other  short  side  will   be  the  altitude  of  the 
triangle. 

43.  In   an    isosceles   triangle,    the   angles   opposite   the 
equal  sides  are  also  equal.      Thus,  in  Fig.  .'35, 

A  B  =  B  C\  hence,  angle  C  =  angle  A. 

In  any  isosceles  triangle,  if  a  perpendicular 
be  drawn  from  the  vertex  opposite  the  unequal 
side  to  that  side,  it  bisects  (cuts  in  halves) 
the  side.  Thus,  A  C  is  the  unequal  side  in  the 
isosceles  triangle  A  B  C;  hence,  the  perpen- 
dicular B  D  bisects  A  C,  or  A  D  =  D  C. 

If  two  angles  of  any  triangle  are  equal,  the  triangle  is 
isosceles. 

44.  In  any  triangle,  the  sum  of  the  three  angles  is  equal 
to  two  right  angles.    Thus,  in  Fig.  o'i,  the  sum  of  the  angles  at 

A,  />,  and  C  =  two  right  angles;  that 
is,  A -\- B -}- C  =  two  right  angles. 
Hence,  if  any  two  angles  of  a  triangle 

L ^     are  given,  the  third  may  be  found  by 

FIG.  36.  subtracting  the  sum  of  the  two  from 

two  right  angles.      Suppose  that  A  -\-  B  =  1  T7()  right  angles ; 
then,  C  must  equal  2  —  1T"7  =  ^  of  a  right  angle. 

'  45.  In  any  right-angled  triangle  there  can  be  but  one 
right  angle,  and  since  the  sum  of  all  the  angles  equals  two 
right  angles,  it  is  evident  that  the  sum 
of  the  two  acute  angles  must  be  equal 
to  a  right  angle.  Therefore,  if  in  any 
right-angled  triangle  one  acute  angle  is 
known,  the  other  can  be  found  by  sub- 
tracting- the  known  angle  from  a  right 
angle.  Thus  ABC,  Fig.  37,  is  a  right- 
angled  triangle,  right-angled  at  C.  Then,  the  angles 
A+B  =  one  right  angle.  If  A  =  f  of  a  right  angle,  B 
=  I  —  T-  =  7  of  a  right  angle. 

46.     In  any  right-angled  triangle,  the  square  described 
on   the   hypotenuse    is  equal   to   the   sum    of   the   squares 


10      GEOMETRY  AND  MENSURATION.     §  4 

described  upon  the  other  two  sides.     If  A  B  C,  Fig.  38,  is  a 

right-angled    triangle,    right- 
angled  at  B,  then  the  square 
I  /N/^x  described   upon    the    hvpote- 

/*"•.  /   *+"   i  -1.       /     *"•>>.  J   ± 

A  C  is  equal  to  the  sum 
squares  described  upon 
sides  A  B  and  B  C;   con- 
.  sequently,    if   the    lengths   of 

4U.[*-!-?i*l-*Jc        the  sides  AB  and  BC  are 

1  6  i  7  j «  i  9  \  10  \  known,  we  can  find  the  length 

of  the  hypotenuse  by  adding 
the  squares  of  the  lengths  of 

\2i\22  \23\24\25\  the   sides  AB  and  B  C,   and 

D  ""  E 

Fic  gg  then    extracting    the    square 

root  of  the  sum. 

If  the  length  of  one  of  the  short  sides  be  denoted  by  #, 
that  of  the  other  short  side  by  b,  and  that  of  the  hypotenuse 
by  c,  then,  since  c~  =  a'-\-b^,  we  have,  by  extracting  the 
square  root  of  both  quantities, 

EXAMPLE. — The  width  of  a  house  is  20  feet,  and  the  roof  is  half 
pitch — that  is,  the  height  of  the  gable  is  equal  to  one-half  the  -width. 
What  is  the  length  of  a  rafter,  if  it  projects  15  inches  over  the  side  ? 

SOLUTION. — Let  one-half  the  width  be  called  a  —  10  feet;  the  height 
of  the  gable,  b  =  \  of  20,  or  10  feet ;  and  the  hypotenuse,  c.  Substi- 
tuting these  values  in  the  formula, 


c  =  tfa*  +  b-  =  |/102  +  102  =  -v/SOO  =  14.14  feet. 

Adding  15  inches  =  1.25  feet,  the  length  of  a  rafter  is  15.39  feet;  or, 
reducing  .39  foot  to  inches,  the  length  is  15  feet  4|  inches,  nearly.      Ans. 

47.  If  the  hypotenuse  and  one  side  are  given,  the  other 
side  can  be  found  by  subtracting  the  square  of  the  given 
side  from  the  square  of  the  hypotenuse,  and  then  extract- 
ing the  square  root  of  the  remainder.  That  is, 

a  =  Vc*  -  P  or  b  =  Vc*-a\ 
the  letters  having  the  same  meaning  as  in  Art.  46. 

EXAMPLE. — It  is  desired  to  ascertain  the  height  of  the  ceiling  in  a 
room.  One  end  of  a  10-foot  pole  is  set  in  the  corner  of  the  ceiling  and 


GEOMETRY  AND  MENSURATION. 


wall,  and  the  other  end  is  0  feet  from  the  base  of  the  wall.     What  is 
the  height  of  the  ceiling  ? 

SOLUTION. — In  this  problem,  the  10-foot  pole  is  the  hypotenuse  c,  and 
the  short  side  b  is  6  feet.     Hence,  applying  the  formula, 


a  =  4/6--  -  //-  :=  4/100  -  36  =  8  feet, 
which  is  the  height  of  the  ceiling.     Ans. 

48.  If  the  sides  are  equal,  or  if,  in  Fig.  30,  a  =  b,  then 
the  hypotenuse  is  equal  to  the  square  root  of  twice  the 
square  of  either  side;  that 
is,  c  =  \/r^li-  =  4/2T2.  Also, 

a  —  b  —  i/~  I  that  is,  either 

'        '/i 

side  is  equal  to  the  square 
root  of  one -half  the  sqiiare  of 
the  hypotenuse. 

In  such  a  triangle,  the  per- 
pendicular distance  from  the  hypotenuse  to  the  right  angle  is 
one-half  the  hypotenuse.  Thus,  in  roofing,  if  G  D,  Fig.  30, 
is  equal  to  £  G,  or  i  E  F,  then  the  roof  is  called  half 'fitch. 

EXAMPLE. — What  is  the  length  of  a  slope  of  a  roof  having  half  pitch, 
the  width  of  the  house  being  30  feet  ? 

SOLUTION. — In  this  example,  c  =  30  feet,  and  a  =  b.     Then,  by  the 
formula, 

a  =  b  =  |/£y.  =  4/4~50  =  21.21  feet.     Ans. 

EXAMPLE. — It  is  desired  to  lay  out  a  line  A  C  at  right  angles  to  a  line 
A  J5,  which  is  15  feet  long.     How  can  it  be  done  by  means  of  a  tape  ? 
SOLUTION. — If  we  can  find   any  two  numbers,  the  sum  of   whose 
squares  is  equal  to  the  square  of  another 
number,  these  three  numbers  will  be  the 
sides  of  a  right  triangle ;  as,  for  exam- 
ple, 3,  4,  and  5,  or  6,  8,  and  10.     Thus, 
hold  the  6-foot  mark  on  the  tape  at  A, 
Fig.  40,  the  14-foot  mark  at  D,  and  bring 

B  the  end  of  the  tape  and  the  24-foot  mark 

together  at  C,  and  mark  the  point  C,  then 
A  C  will  be  perpendicular  to  A  D ;  for 
tfA  C3  +  A  D*  =  CZ>;  that  is,  4/6" +  8"  =  4/36  +  64  =  10  feet  =  24  feet 
- 14  feet. 


FIG.  40. 


12 


GEOMETRY  AND  MENSURATION. 


49."  The  principle  of  the  right  triangle  is  of  very  great 
value  in  practical  work,  and  the  student  should  become 
thoroughly  familiar  with  it  in  all  its  variations. 

The  following  is  an  example  showing  a  double  application 
of  the  right-triangle  principle : 

EXAMPLE. — In  Fig.  41,  A  £  CD  represents  a  skylight  7  ft.  6  in.  X  9 

ft.  The  point  O  is  2  feet  above 
the  plane  of  A  B  CD.  What 
is  the  length  of  the  hip  rafter 
O  D  at  the  angle  of  the  skylight? 

SOLUTION. — It  will  be  seen 
that  the  point  O'  is  directly 
under  O  and  2  feet  below  it,  so 
that  O'  is  on  the  same  level  as 
A  B  CD.  It  will  also  be  seen 
that  D  O  is  simply  the  hypote- 
nuse of  a  right  triangle,  whose 
base  is  D  O'  and  whose  altitude 
is  O  O'.  But  D  O'  is  also  the 
hypotenuse  of  a  right  triangle, 
whose  sides  are  EO'  and  ED. 
EO'  =  \  of  7  feet  6  inches 
=  3.75  feet,  and  ED  =  \  of  9 
feet  =  4.5  feet.  Then,  DO'  =  t/(3.75)2  +  (4.5)2  =  5.86  feet.  In  the 
triangle  DO'  O,  DO'  =  5.86  feet,  and  O  O'  =  2  feet;  whence, 


FIG.  41. 


DO  —  \/(5.8Q)'2  +  2'2  =  6.19  feet  =  6  feet  2£  inches,  nearly. 
One  extraction  of  square  root  may  be  dispensed  with,  thus : 
DO  = 


=  V(8.75)s 


'Y  +  (EDY]  +  (00'Y 
=  ^38.125  =  6.19  feet,  as  before.     Ans. 


Again,  if  we  can  find  the  length  of  a  line  along  the  skylight  perpen- 
dicular to  an  edge,  as  O  E  to  AD,  then  the  hip  O  D  is  the  hypot- 
enuse of  a  right  triangle  of  which  O  E  and  E  D  are  sides  ;  and  O  E 


The  result  will  be  the  same  either  way. 

NOTE.  —  When  decimals  occur  in  the  answers  to  the  examples  in  this 
section  on  Geometry  and  Mensuration,  two  decimal  places  are  to  be 
retained,  and  if  the  third  decimal  figure  is  greater  than  5,  the  second 
decimal  figure  is  to  be  increased  by  1;  thus,  13.537  should  be  written 
13.54,  not  13.53.  To  easily  convert  feet  and  inches  to  feet  and  decimals 
of  a  foot,  feet  and  decimals  of  a  foot  to  feet  and  inches,  or  to  change  a 
fraction  of  an  inch  to  a  decimal,  or  vice  versa,  the  conversion  tables, 
Art.  82,  may  be  used. 


GEOMETRY  AND  MENSURATION. 


13 


SIMILAR  TRIAXfJ  LTCS. 

50.  Two  triangles  arc   equal  when  the  sides  of  one  are 
equal  to  the  sides  of  the  other. 

51.  Two  triangles  are  similar  when  the  angles  of  one 
are  equal  to  the  angles  of  the  other.      The  corresponding 
sides  of  similar  triangles  are  proportional. 

For  example,  suppose  we  have  two  triangles  A  B  C  and 
a  be,  Fig.  42,  in  which  the  side  ac 
is  perpendicular  to  A  C,  the  side 
a  6,  to  A  />,  and  side  c  b,  to  B  C, 
then,  angle  A  =  angle  a,  since  the  A 
sides  of  one  are  perpendicular  to  the 
sides  of  the  other.  (See  Art.  £4.) 
In  like  manner,  angle  B  =  angle 
$,  and  angle  C  =  angle  e.  The 
two  triangles  are  therefore  similar, 
and  their  corresponding  sides  are 
proportional.  That  is,  any  two 
sides  of  one  triangle  are  to  each  other  as  the  two  corre- 
sponding sides  of  the  other  triangle ;  or,  one  side  of  one  tri- 
angle is  to  the  corresponding  side  of  the  other  as  another  side 
of  the  first  triangle  is  to  the  corresponding  side  of  the  second. 
The  following  are  examples  of  the  many  proportions  that 
may  be  written.  In  this  case,  the  corresponding  sides  of 
the  two  triangles  are  the  ones  perpendicular  to  each  other. 

AB:BC  =    ab:bc,          BC'.bc     -AB-.ab, 
AB\AC  —    a  b  :  a  e,          A  C:  a  c  =  B  C  :  b  c,  etc. 
EXAMPLE. — It  is  required  to  find  the  distance  A  />',  Fig.  4:5,  across  a 
stream. 

SOLUTION. — The  line  B  C  making  any 
angle  with  A  B  is  measured,  and  C  E  is 
made  parallel  with  A  7>,  and  of  any  con- 
venient length.  The  point  D  is  marked 
where  A  E  intersects  />  C,  and  B  D  ami 
D  C  are  measured.  Then  since  the  tri- 
angle AB  D  and  CD  E  have  their  cor- 
responding sides  parallel,  they  are  simi- 
lar, and  AB:CE  =  BD-.DC;  or  AB 
30X96 


FIG.  43. 


20 


=  144  feet.     Ans. 


14 


GEOMETRY  AND  MENSURATION. 


If  a  straight  line  be  drawn  through  two  sides  of  a  tri- 
angle parallel  to  the  third  side,  it  divides 
those  sides  proportionally.  Thus,  let  the 
line  D  E  be  drawn  parallel  to  the  side  B  C 
in  the  triangle  ABC,  Fig.  44.  Then, 
AD-.AB  =  AE-.AC,  or  A  D  :  D  B 
=  A  E  :  E  C.  It  is  to  be  noticed  that 
the  triangles  A  D  E  and  ABC  are 
similar,  and  their  sides  are  propor- 
tional. Thus  A  B  \  A  D  =  B  C\  D  E  and 
A  C:AE  =  BC-.DE. 
If  a  straight  line,  as  D  F,  be  drawn  from  D  or  E,  Fig.  44, 

parallel  to  A  C  or  A  B,  then  the 

triangles     A  D  E,     ABC,    and 

DBF   are    all    similar,    and    a 

number    of    other    proportions 

may  be  formed. 

EXAMPLE. — Referring  to  Fig.  45,  it  is 
desired  to  find  the  length  of  the  line 
CA,  extending  across  a  river.  E  is  on 
the  line  A  C,  D  is  on  the  line  B  A, 
and  D  E  is  parallel  to  C  B ;  the  lengths 
of  CE,  E  D,  and  C  B  are  as  shown  in 
the  figure. 

SOLUTION. — Draw  EF  parallel  to 
A  B,  so  that  CF  =  125  —  90  =  35 feet; 

then  CA:CE=  CB-.CF,  or  CA : 60  =  125:35, whence,  CA  = 
=  214. 3  feet,  nearly. 

EXAMPLE. — On  a  drawing  it  is  required  to  divide  a  line  8  inches  long 
•  into  12  parts. 

SOLUTION.— Let  A  B,  Fig.  46,  be  the 
8-inch  line.  Through  A  draw  any  line 
A  C  12  inches  long,  and  mark  the  inch 
points  on  it.  Connect  C  and  B,  and 
through  each  inch  mark  on  A  C,  draw 
a  parallel  to  CB,  as  D  E,  etc.,  cutting  A  B 
at  E,  etc.,  which  will  be  the  points  re- 
quired ;  for  by  similar  triangles,  A  E:  A  D 
1X8 


FIG.  45. 


60  X  125 
35 


FIG.  46. 


=  AB:A  C;or  A E:\  in.  =  8 in.:  12 in. ;  hence,  A  E  — 


12 


=  |  inch. 
Ans. 


GEOMETRY  AND  MENSURATION. 


15 


This  principle  is  very  useful  when  an  exaet  measurement — 
as  for  example,  f  inch — cannot  be  obtained  by  a  scale  or 
rule. 


EXAMPLES  FOR  PRACTICE. 
53.      Solve  the  following: 

1.  The  distance  from  the  first  to  the  second  floor  of  a  house  is  9  feet. 
It  is  desired  to  mark  the  line  of  the  top  of  the  14  steps,  or  treads,  on  a 
drawing.     What  is  the  height  of  each  riser  ?     Make  a  sketch  showing 
how  the  spacing  may  be  found.  Ans.   7.7  in.   =  ~i\\  in.,  nearly. 

2.  What  length  of  stone  coping  is  required  for  each  side  of  a  gable, 
the  width  of  which  is  24  feet,  and  whose  height  is  -J  the  span  ? 

Ans.   21.03  ft. 

3.  What  is  the  angle  at  the  top  of  a  gable  of  a  house  whose  roof 
slopes  are  each  one-half  of  a  right  angle  ?  Ans.   A  right  angle. 

4.  In  running  a  line  FA,  Fig.  47,  a  house  stands 
in  the  way.     A  C  is  laid  off  square  with  A  F,  14  feet 
long,  and  12  feet  behind  A  C,  a  line  G  E  23  feet  long 
is  laid  off  at  right  angles  to  A  F,  so  as  to  give  a  sight 
past  the  house  to  B.     What  are  the  lengths  of  A  B 
and  CB1     CE  =  |/9*  +  12*.  (  A  B,  18-f  ft. 

nS"  '(   CB,  23L  ft. 

5.  What  is  the  distance  between  the  16-inch  mark 
on  one  leg  of  a  carpenter's   square   and  the   12-inch 
mark  on  the  other  ?  Ans.  20  in. 

4  6.     The  distance  EB,  Fig.  4S, 

is  8'-  feet.  If  the  spacing  of  the 
roof  rafters  is  20  inches,  and  the 
true  length  of  A  B  is  12  feet,  what 
is  the  length  of  the  "  jack  rafter" 
CD1 
Ans.  7|  ft.  =  7ft.  2\  in.,  nearly. 

7.  In  Fig.  49,  C  B  A  EC  represents  a  gable  at  right  angles  to  the 
main  roof,  the  line  A  B,  where  they  meet,  being  called  a  valley.  What 
is  the  length  of  the  valley 
rafter  A  B,  C  A  being  16  feet, 
ED,  the  rise  of  the  rafter, 
being  8  feet,  andEfi  =  D  F, 
the  distance  of  B  back  of 
A  C  E,  4  feet  ?  The  principle 
is  the  same  as  explained  for 
hips.  Ans.  12  ft.  Fir..  49. 

1-10 


B 

FIG.  48. 


16 


GEOMETRY  AND  MENSURATION. 


THE    CIRCLE. 

54.  A  circle  is  a  plane  figure  bounded 
by  a  curved  line,  called  the  circumference, 
every  point  of  which  is  equally  distant  from 
a  point  within,  called  the  center.  See  Fig. 
50. 


FIG.  50. 


55.     The   diameter   of   a    circle   is   a 

straight  line  passing  through   the  center  A[ , ]g 

and  terminated  at  both  ends  by  the  cir-  \  / 

cumference,  as  A  B,  Fig.  51. 

FIG.  51. 

,,-- — -v,  56.     The  radius  of  a  circle,  OA,  Fig. 

/  \  52,  is  a  straight  line  drawn  from  the  center 

\  to  the  circumference.     It  is  equal  in  length 

/  to  one-half  the   diameter.      The  plural  of 

*/'  the  word  radius  is  radii.     All  radii  of  any 

FIG  K  circle  are  equal  in  length. 


57.     An  arc  of  a  circle  is  any  part  of  its 
circumference,  as  A  E  B,  Fig.  53. 


FIG.  53. 


58.  A  chord  is  a  straight  line  joining 
any  two  points  in  a  circumference ;  or,  it  is 
a  straight  line  joining  the  extremities  of  an 
arc.  Thus,  in  Fig.  54,  A  B  is  the  chord  of 
the  arc  A  E  B. 


59.  A  segment  of  a  circle  is  the  space  included 
between  the  arc  and  its  chord.  Thus,  in  Fig.  54,  the  space 
between  the  arc  A  E  B  and  the  chord  A  B  is  a  segment. 


GEOMETRY  AND  MENSURATION. 


17 


GO.     A  sector  of  a  circle  is  the  space 
included    between    an    are    and    two    radii    A\ 
drawn    to   the   extremities   of    the   arc,   as 
A  OB,  Fig.  55. 


61.  Two  circles  are  equal  when  the  radius  or  diam- 
eter of  one  is  equal  to  the  radius  or  diameter  of  the 
other. 

Two  arcs  are  equal  when  the  radius  and  cJiord  of  one  is 
equal  to  the  radius  and  chord  of  the  other. 


62.  If  A  D  B  C,  Fig.  50,  is  a  circle 
in  which  two  diameters  A  B  and  CD 
are  drawn  at  right  angles  to  each  other, 
then  A  O  D,  DOB,  B  O  C,  and  CO  A 
are  right  angles.  The  circumference 
is  thus  divided  into  four  equal  parts; 
each  of  these  parts  is  called  a  quml- 
rant. 


63.  To  measure  angles,  the  circumference  of  circles  are 
divided  into  3GO  equal  parts  called  degrees,  which  are  sub- 
divided into  GO  equal  parts  called  minutes;  and  the  latter 
are  further  subdivided  into  GO  equal  parts  called  seconds. 
Degrees,  minutes,  and  seconds  are  indicated  by  the  marks 
°,  ',  ";  thus,  G5  degrees,  15  minutes,  and  40  seconds  is 
written  65°  15'  40".  Since  a  quadrant  is  one-fourth  of  a  cir- 
cumference, it  includes  \  of  3GO°,  or  00°,  whence  a  right 
angle  contains  90°.  So,  also,  if  a  circle  be  divided  into 
equal  sectors,  the  angle  included  between  two  adjacent 
radii  is  equal  to  3GO°  divided  by  the  number  of  sectors. 
Thus,  the  angle  between  radii  drawn  to  the  angles  of  a 
regular  octagon  includes  ^  of  360°,  or  45°. 

The  intersection  of  any  two  straight  lines  may  be  con- 
sidered the  center  of  a  circle,  and  the  number  of  3GOths,  or 
degrees — measured  on  any  circumference  described  from 


18 


GEOMETRY  AND  MENSURATION. 


this    center — included    between    the    lines,    measures    the 
angle. 

64.     An  inscribed  angle  is  one  whose  vertex  lies  on  the 
circumference  of  a  circle,  and  whose  sides  are  chords.     It  is 
B  measured  by  one-half 'the  intercepted  arc. 

Thus,  in  Fig.  57,  ABC  is  an  inscribed 
angle,  and  it  is  measured  by  one-half  the 
arc  ADC. 


EXAMPLE.— If,  in  Fig.  57,  the  arc  A  D  C  =  f  of 
YC  the  circumference,  how  many  degrees  are  there 
in  the  inscribed  angle  A  B  C  ? 

SOLUTION. — Since  the  angle  is  an  inscribed 
angle,  it  is  measured  by  one-half  the  intercepted 
arc,  or  f  X  *  =  i  °f  the  circumference.  The  whole  circumference  con- 
tains 360° ;  and  360°  X  \  =  72°.  Ans. 

65.     If  a  circle  is  divided  into  halves,  each  half  is  called 
a  semicircle,  and  each  half  circumfer- 
ence is  called  a  semi-circumference. 

The  measure  of  a  semicircle  is  one-half 
of  360°,  or  180°. 

Any  angle  that  is  described  in  a  semi- 
circle and  intercepts  a  semi-circumfer- 
ence, as  A  B  C  or  A  D  C,  Fig.  58,  is 
a  right  angle,  since  it  is  measured 
by  one-half  a  semi-circumference,  or  by  90C 


FIG.  58. 


66.     If,  in  any  circle,  a  radius  be  drawn  perpendicular 
to  any  chord,  it  bisects  (cuts  in  halves)  the  chord.     Thus, 
if  the  radius   O  C,   Fig.  59,  is  perpendic- 
ular to  the  chord  A  B,  A  D  —  D  B. 

A  radius  which  bisects  a  chord,  bisects 
also  the  angle  included  between  radii 
drawn  to  the  ends  of  the  chord.  Thus, 
in  Fig.  59,  the  radius  O  C  bisects  the  angle 
A  OB. 

If  a  straight  line  be  drawn  perpendic- 
ular to  any  chord  at  its  middle  point,  it  must  pass  through 
the  center  of  the  circle. 


FIG.  59. 


GEOMETRY  AND  MENSURATION. 


19 


67.  Through  any  three  points  not  in  the  same  straight 
line,  a  circumference  can  be  drawn.  Let 
A,  £,  and  C,  Fig.  GO,  be  any  three  points. 
Join  A  and  />',  and  B  and  C  by  straight 
lines.  At  the  middle  point  of  A  />,  draw 
H K  perpendicular  to  A  B;  at  the  middle 
point  of  B  C  drawr  E  F  perpendicular  to 
B  C.  These  two  perpendiculars  intersect 
at  O.  With  (9  as  a  center,  and  OB,  O  A, 
or  O  C  as  a  radius,  describe  a  circle ;  it  will  pass  through 
A,  B,  and  C. 


68.  A  tangent  to  a  circle  is  a  straight 
line  which  touches  the  circle  at  one  point 
only;  it  is  always  perpendicular  to  a 
radius  drawn  to  that  point.  Thus,  A  />, 
Fig.  61,  is  a  tangent  to  the  circle;  it 
touches  the  circle  at  E  and  is  perpendic- 
ular to  the  radius  O  E. 


FIG.  ei. 


69.  If  two  circles  intersect 
each  other,  the  line  joining  their 
centers  bisects  at  right  angles  the 
line  joining  the  two  points  of  inter- 
section. Thus,  if  the  two  circles, 
whose  centers  are  O  and  P,  Fig. 
62,  intersect  at  A  and  B,  the  line 
O  P  bisects  at  right  angles  the  line 
AB;  or  A  C  =  BC. 


FIG.  63. 


7O.  One  circle  is  said  to  be 
tangent  to  another  circle  when 
they  touch  each  other  at  one  point 
only.  See  Fig.  03.  This  point  is 
called  the  point  of  contact,  or 
the  point  of  tangency. 


20 


GEOMETRY  AND  MENSURATION. 


When  two  or  more  circles  are  described  from  the 
same  center,  they  are  called  concentric 
circles.  See  Fig.  64. 


FIG.  64. 


H 


FIG.  65. 


72.  If,  from  any  point  on  the  circumfer- 
ence of  a  circle,  a  perpendicular  be  let  fall 
upon  a  given  diameter,  this  perpendicular 
will  be  a  mean  proportional  between  the  two 
parts  into  which  it  divides  the  diameter. 
If  A  B,  Fig.  65,  is  the  diameter,  and  C  any  point  on  the 
circumference,  then  the  perpendicular  CD  is  a  mean  pro- 
portional between  A  D  and  D  B,  or 
AD:  CD  =  CD\DB.  Therefore,"^2 
=  ADxDB,mACD  =  VADxDB. 
This  principle  furnishes  a  method  of 
drawing  a  mean  proportional  between 
two  lines.  Let  A  D  and  D  B,  Fig.  65, 
be  any  two  lines.  Join  them  together 
in  one  line  as  A  £>,  and  on  this  as  a 
diameter,  draw  a  circle.  Then  CD,  perpendicular  to  this 
diameter  at  the  common  end  D,  is  the  mean  proportional. 

EXAMPLE. — In  arches  the  span  is  the  distance  across  the  opening,  as 
A  C,  Fig.  66,  measured  from  the  ends  of  the  arch  as  A  and  C.     The 

rise  D  B  is  the  perpendicular  distance 
from  A  C  to  the  highest  point  B  meas- 
ured on  the  center  line  OB ;  D  B  —  the 
radius  —  OD. 

The  span  A  C,  Fig.  66,  of  an  arch  is 
6  feet,  and  the  rise  D  B  is  8  inches. 
What  is  the  length  of  the  radius  O  Bl 

SOLUTION.  —  Here  the  span  is  the 
chord  of  a  circle,  and  since  a  radius  per- 
pendicular to  a  chord  bisects  it,  A  D  is 
|  of  A  C  =  36  inches.  If  we  draw  the 
complete  circle,  it  will  be  seen  that  ED 
and  D  B  are  the  parts  of  the  diameter 
and  A  D  the  perpendicular  from  the 
point  A.  Then  ^-J?  =  EDY.DB, 
or  36"  =  ED  X  8,  whence,  ED  =  1,296 
-5-  8  =  162  inches,  and  EB  =  162  +  8  =  170  inches.  OB  =  £  EB  =  85 
inches,  or  7  feet  1  inch.  Ans. 


GEOMETRY  AND  MENSURATION. 


INSCRIBED    AND    CIRCrMSCRIBKD    POLYGONS. 

73.  An    inscribed  polygon    is    one  whose   vertices    lie 
on   the   circumference   of    a  circle 

and    whose    sides    are    chords,    as 
KLMNPQ,  Fig.  U7. 

74.  A  circumscribed  polygon 


FIG.  <;r. 


is  one  whose  sides  are  tangent  to  a 
circle,  a&ABCDEF,  Fig.  07. 

Circles  may  be  inscribed  in  and 
circumscribed  about  any  regular 
polygon.  The  radius  of  a  circle  is 
the  distance  from  the  center  to  an  angle  of  the  inscribed 
polygon,  and  the  perpendicular  distance  from  the  center 
to  the  side  of  the  circumscribed  polygon,  as  shown  in 
Fig.  67. 

75.  If  lines  be  drawn  from  the  center  to  the  angles  of 
a  regular  polygon,  they  will  make  equal  angles  with  the 
sides  of  the  polygon  and  will  form  isosceles  triangles.  As 
the  sum  of  all  the  angles  formed  by  lines  drawn  from  a 
point  is  4  right  angles,  or  300°,  the  angle  at  the  center  of  a 
regular  polygon  between  two  radii  equals  300°  divided  by 
the  number  of  sides.  Thus  the  angle  between  two  radii 
drawn  to  the  end  of  a  side  of  a  regular  hexagon  is  360°  -f-  6 
=  60°.  The  sum  of  the  angles  of  a  triangle  is  2  right 
angles,  or  180° ;  hence,  since  the  triangles  formed  by  draw- 
ing radii  to  the  angles  of  a  regular  polygon  are  isosceles, 
the  angles  OAB  and  O B A,  Fig.  G7,  are  each  equal 
to  \  the  difference  between  180°  and  the  central  angle. 
Thus,  each  angle  =  ix(180°  —  00°)  =  60°,  and  the  triangle 
OB  A  is  equilateral.  In  a  regular  hexagon,  therefore, 
the  sides  are  equal  to  the  radius  of  the  circumscribed 
circle. 

A  line  drawn  from  the  center  of  a  regular  polygon  to 
an  angle  bisects  the  angle;  thus  OA,  Fig.  67,  bisects  the 
angle  B  A  F,  which,  therefore,  is  equal  to  %xOA£,  or 
2XOAF. 


22 


GEOMETRY  AND  MENSURATION. 


FIG. 


EXAMPLE. — As  the  principles  of  Art.  75  are  used  in  forming  bevel 
angles,  etc.,  let  us  find  at  what  angle  the 
bevel  must  be  set  to  mark  the  cuts  for 
the  moldings  around  an  octagonal  room 
of  equal  sides,  as  in  Fig.  68. 

SOLUTION. — Draw  AO,  BO,  etc.  to  the 
center  O,  forming  triangles  A  O  B,  B  O  C, 
etc.  Then  each  of  the  angles  at  O  is  £  of 
360°  =  45°.  The  angles  O  A  B  and  O  B  A 
are  each  \  X  (180°  -45°) •=  67i°.  If  the 
blade  of  the  bevel  is  set  at  this  angle,  as 
at  D,  and  the  ends  of  the  pieces  of  molding 
are  cut  to  it,  they  will  fit  together  properly. 

EXAMPLE. — In  Fig.  69,  G  H E F 'and  H K D  E  represent  two  boards 
forming  a  center  for  supporting  a  brick 
arch  during  building ;  the  shaded  parts 
being  cut  off,  leaving  a  curve,  as 
ALB.  What  is  the  length  of  each 
board,  and  the  angle  of  bevel  G  H  O 
where  they  meet  ?  The  radius  A  O  of 
the  arch  is  2  feet. 

SOLUTION. — Draw  G  H  parallel  to 
A  B  and  meeting  O  G  and  O  H,  and 
also  O  L  perpendicular  to  A  B;  then 
A  M  =  MB  (Art.  66),  and  by  similar 
triangles,  L  G  =  L  H.  O  L  bisects  the  right  angle  O  A  B,  hence, 
A  O  M  =  |  of  90°,  or  45°.  Then  the  other  acute  angle  in  the  right 
triangle  O  M  A  =  90°  —  45°  =  45°  also.  The  angle  L  G  O  is  equal  to 
angle  M  A  D,  or  45°,  and  the  right  triangle  O  L  G,  having  two  angles 
equal,  the  opposite  sides  are  equal  also,  and  since  O  L  —  the  radius, 
or  2  feet,  L  G  is  also  2  feet  and  GJ/7=2X2  =  4  feet,  the  required 
length.  Angle  O  H  G  =  angle  O  G  H  =  45°,  the  angle  a't  which  to 
set  the  bevel. 


EXAMPLES  FOR  PRACTICE. 

76.   .  Solve  the  following: 

1.  What  angle  does  the  minute  hand  of  a  clock  travel  over  in  5 
minutes  ?  Ans.  30°. 

2.  The  span  of  an  arch  is  7  feet,  and  the  rise  is  1  inch  for  each  foot 
of  span.     What  is  the  radius  of  the  arch  ?  Ans.   10  ft.  9£  in. 

3.  The  centering  for  an  arch,  Fig.  70,  8  feet  in  diameter  is  made 
of  3  pieces  of  equal   length.     Knowing   that  the    side  of    a    regular 


§4  GEOMETRY  AND  MENSURATION.  #j 

inscribed  hexagon  is  equal  to  the  radius,  what  must  be  the  length,  as 
A  J3,  of  each  piece  ? 

Ans.  4.62  ft.,  or  4  ft.  7i  in.,  nearly. 
Query. — What   is   the   length   of    O  C  in 
right  triangle  O  CJi! 

4.  In  the  preceding  example,  what  is  (a) 
the  bevel  angle,  and  (b)  the  angle  between 
any  two  pieces  ? 


5.  The  flooring  in  a  regular  octagonal  room, 
Fig.  71,   is    laid    parallel   with    the    side   ./  />', 
which  is  (5  feet  long.     What  are  the  angles  of 
bevel  for  cutting  the  ends  along  .  /  C'and  CD  ? 

Ans.   67 T  and  90°. 

6.  What    is    the    length    of    the    flooring 
between  CD  and  //A',  Fig.  71,  knowing  that 
C/'and  A  F  are  equal,  and  that  angle  C  l-\[ 
is  a  right  angle  ? 

Ans.   14.48  ft.  or  14  ft.  5f  in.,  nearly. 


ME^SUKATIOX. 


INTRODUCTION. 

77.  Mensuration  is  that  part  of  geometry  which  treats 
of  the  measurement  of  lines,  surfaces,  and  solids. 

78.  The   practical    application    of    mensuration    is   the 
computation   of  lengths,   areas  of  surfaces,   or  volumes  of 
solids.     The  dimensions   which   furnish  the  data  required 
are   usually    obtained    from    working   drawings,    or   plans; 
therefore,  before  formally  taking  up  the  subject  of  mensu- 
ration, it  is  necessary  to  explain  how  drawings  and  plans 
are  made,  and  how  the  dimensions  are  taken  from  them. 

79.  Working   drawings   are   generally   made    to   scale; 
that  is,  the  lines  on  the  drawing  have  a  certain  ratio  to  the 
corresponding  dimension  of  the  full-size  object  which  the 
drawing  represents;    this   ratio   is   called    the    scale.      For 
example, .if  the  scale  of  a  drawing  is  \  inch  to  1  foot,  it 


24      GEOMETRY  AND  MENSURATION.     §  4 

means  that  each  half  inch  on  the  drawing  represents  1  foot 
on  the  object,  and  the  drawing  is  ^-f-12  =  Tx¥  of  full  size. 

8O.  To  make  drawings  and  to  obtain  measurements 
from  them,  divided  rules  called  scales  are  used.  These  con- 
sist of  strips  or  pieces  of  wood  or  metal  1  or  2  feet  long, 
having  the  edges-  beveled,  and  upon  which  are  engraved 
the  graduations  for  the  different  scales,  as  1  inch  to  1  foot, 
\  inch  to  1  foot,  etc.  The  marks  are  numbered  in  order 
from  the  end ;  the  space  next  the  end  being  subdivided  into 
twelfths,  representing  inches,  and  in  the  larger  scales,  these 
spaces  are  again  subdivided  for  fractions  of  an  inch.  There 
are  usually  two  scales  marked  on  each  edge,  and  in  one  the 
divisions  are  twice  as  long  as  in  the  other,  and  are  numbered 
from  the  other  end. 

To  measure  a  line  on  a  drawing,  place  the  scale  with  the 
proper  edge  along  the  line,  and  move  it  until  one  of  the  foot 
marks  is  opposite  one  end  of  the  line,  as  at  b,  Fig.  72,  while 


*ui 

1  11"1 

1 

,'''!'  T  '', 

1  1  ' 
. 

•  i  ' 

< 

1  1  1 

:             i 

1    1   ' 

• 

/"]"'. 

."T",1;",* 

FIG.  72. 

the  other  end  of  the  line  is  at  the  0  mark  on  the  scale,  or 
opposite  one  of  the  small  graduations,  as  at  a.  Then  read 
the  number  of  large  divisions  from  the  0  mark  to  the  far 
end  of  the  line,  and  also  the  number  of  small  divisions  from 
0  to  the  near  end  of  the  line.  Thus  in  Fig.  72,  the  line  a  b 
extends  over  9  large  spaces  and  5  small  ones,  or  -^  of  a  large 
one,  so  that  the  length  of  the  line  represented  by  a  b  is  9  feet 
5  inches. 

81.  Working  drawings  usually  have  the  principal  dimen- 
sions marked  on  them,  as  shown  in  Fig.  73,^vhich  represents 
a  side  and  an  end  view  of  a  cast-iron  lintel.  The  length  is 
7  feet  6  inches,  as  indicated  by  the  broken  line  c  d  extending 
between  ec  and  fd,  and  drawn  perpendicular  to  ef  at  the 
ends  e  and  f.  The  arrowheads  indicate  the  ends  of  the 


GEOMETRY  AND  MENSURATION. 


7' 6"  space.  At  //  is  shown  that  the  height  from  ^  to  A' is 
6  inches,  being  the  distance  between  the  parallel  lines  ;//  A" 
and  n  g.  At  /  are  shown  two  arrowheads  pointed  towards 
each  other;  these  are  used  where  the  space  is  too  small  to 


insert  the  arrowheads  and  figures  in  the  ordinary  manner. 
In  this  case  it  means  that  the  thickness  of  the  rib  is  '\  inch, 
as  shown  by  the  nearby  figures. 

Having  read  or  scaled  all  the  necessary  dimensions  from 
the  drawing,  the  proper  formulas  or  rules  may  be  applied 
to  obtain  the  required  length,  area,  or  contents. 

82.  The  following  tables  are  useful  in  changing  inches 
or  fractions  of  an  inch,  to  decimals  of  a  foot  or  inch,  and 
vice  versa: 

COXVEKS10X    TAB1VKS. 


IXC'IIKS  TO   DECIMALS  OF  A   FOOT. 


Inches  or 
Fractions. 

Decimal 
of  Foot. 

Approxi- 
mate 
Decimal. 

Inches. 

Decimal 
of  Foot. 

Approxi- 
mate- 
Decimal. 

TV 

.0052 

.005 

3 

.2500 

.25 

i 

.0104 

.010 

4 

.3333 

.33 

1 

.0208 

.020 

5 

.4167 

.42 

3 

¥ 

.0313 

.030 

6 

.5000 

.50 

\ 

.0417 

.040 

7 

.5833 

.58 

I 

.0521 

.050 

8 

.6667 

.67 

f 

..0625 

.060 

9 

.  7500 

.75 

.0729 

.070 

10 

.8333 

.83 

1 

.0833 

.080 

11 

.0167 

.92 

2 

.1667 

.170 

12 

1.0000 

1.00 

26      GEOMETRY  AND  MENSURATION.     §  4 

FRACTIONS  OF  AN  INCH  TO  DECIMALS  OF  AN  INCH. 


Fraction. 

Exact 
Decimal. 

Approxi- 
mate        Fraction. 
Decimal. 

Exact 
Decimal. 

Approxi- 
mate 
Decimal. 

A 

.03125 

.03 

yV 

.5625 

.56 

yV 

.06250 

.06 

5 

.6250 

.63 

¥ 

.12500 

.13 

11 

.6875 

.69 

yV 

.18750 

.19 

I 

.7500 

.75 

i 

.25000 

.25 

If 

.8125 

.81 

A 

.31250 

.31 

7 
¥ 

.8750 

.88 

t 

.37500 

.38 

1  a 
IT 

.9375 

.94 

yV 

.43750 

.44 

1       , 

1.0000 

1.00 

* 

.50000 

.50 

EXAMPLE.—    7  ft.  8J  in.  =7.00  ft.  +.67  ft.  +.07  ft.  =  7.74  ft. 

18.19  ft.  =  18  ft.  +  .17  ft.  +.02  ft.  =  18  ft.  2J  in. 

19T9ff  in.  =  19.56  in. ,  approximately. 

13.26  in.  =  13^  in. ,  approximately. 


MENSURATION  OF    PT^ANE    SURFACES. 

83.  The  area  of  a  surface  is  expressed  by  the  number 
of  unit  squares  it  will  contain. 

84.  A  unit  square  is  the  square  having  the  unit  for  its 
side.      For  example,  if  the  unit  is  1  inch,  the  unit  square  is 
the  square  whose  sides  measure  1  inch  in  length,  and  the 
area  would  be  expressed  by  the  number  of  square  inches 
that  the  surface   contains.      If   the  unit  were   1   foot,    the 
unit  square  would   measure  1  foot  on  each  side,  and  the 
area  would  be  the  number  of  square  feet  that  the  surface 
contains,  etc. 

The  square  that  measures  1  inch  on  a  side  is  called  a 
square  Inch,  and  the  one  that  measures  1  foot  on  a  side  is 
called  a  square  foot.  Square  inch  and  square  foot  are 
abbreviated  to  sq.  in.  and  sq.  ft.,  or  to  cT  and  n'. 


§  4  GEOMETRY  AND  MENSURATION.  •_>: 

TIIK   THIANCiLH. 

85.  Rule. —  To  find  tlic  area  of  a  triangle,  multiply  t/ie 
base  by  tlic  altitude  and  divide  the  product  by  2. 

Let  b  =  base ; 

h  =  altitude; 
A  =  area. 

Then,  A=!£ 

/v 

EXAMPLE. — How  many  square  feet  of  1-inch  boards  will  be  needed  for 
the  2  gables  of  a  house  having  a  half-pitch  roof  and  width  of  24  feet  ? 

SOLUTION. — In  half-pitch  roofs  the  rise  equals  one-half  the  span; 
hence,  h  =  \  of  24  =  12  feet,  and  b  =  24  feet.  Applying  the  formula, 
A  =  \bh  =  \ X  24  X  12  —  144  square  feet  for  each  gable,  or  144  X  2 
=  288  square  feet  of  boards  for  the  2  gables.  Ans. 

86.  If  the  triangle  is  a  right -angled  triangle,  one  of  the 
short  sides  may  be  taken  as  the  base,  and  the  other  short 
side  as  the  altitude ;  hence,  t/ic  area  of  a  right-angled  tri- 
angle is  equal  to  one-half  the  product  of  the  two  sliort  sides. 

87.  The  area  of  any  triangle  may  be  found,  when  the 
length  of    each  side  is  known,  by  means  of  the  following 
formula,  in  which  a,  b,  and  c  represent  the  lengths  of  the  sides, 
s  the  half  sum  of  the  lengths,  and  A  the  area  of  the  triangle : 

A  =  Vs  (s  —  a)  (s  —  l>)  (s  —  c),  where  s  = 

EXAMPLE. — How  many  feet  of  6-inch  clapboards,   laid  4}   inches  to 
the  weather,  will  be  required  for  the  gable 
shown  in  Fig.  74  ? 

SOLUTION. — It  is  immaterial  which  side 
is  called  a,  b,  or  c.     Applying  the  formula, 

a  +  b  +  c      28  +  19.8  +  19.8 
s  =  — — p =  —       — - —       —  =r  do.  8,  the 

half  sum ;  taking  b  and  c  as  the  short  sides, 
s  —  a  =  33.8—28  =  5.8  and  s  -  b  and  s  —  c 
are  each  33.8  —  19.8  =  14.  Now  applying  the  formula, 


A  =  ^s  (s  —  a)  (s  —  b)  (s  —  c)  =  |/33.8  X  5-8  X  14  X  14 

=  196+  square  feet. 

A  clapboard  with  4|-inch  exposure  and  1  foot  long,  has  54  square  inches 
exposed;  1  square  foot,  then,  will  require  144-7-54  =  2J  linear  feet  of 
clapboards,  and  196  square  feet  will  require  196  X  2f  =  522j  feet.  Ans. 


28 


GEOMETRY  AND  MENSURATION. 


THE  QUADRILATERAL. 

88.  A  parallelogram  is  a  quadrilateral  whose  opposite 
sides  are  parallel. 

There  are  four  kinds  of  parallelograms:  the  square,  the 
rectangle,  the  rhombus,  and  the  rhomboid. 


89.  A  rectangle  is  a  parallelogram 
whose  angles  are  all  right  angles.  See 
Fig.  75. 


90.     A    square    is    a    rectangle,  all   of 
whose  sides  are  equal.     See  Fig.  76. 


FIG.  7 


91.  A  rhomboid  is  a  parallelo- 
gram whose  opposite  sides  only  are 
equal,  and  whose  angles  are  not 
right  angles.  See  Fig.  77. 


FIG. 


FIG.  78. 


92.  A  rhombus  is  a  parallelo- 
gram having  equal  sides,  and 
whose  angles  are  not  right  angles. 
See  Fig.  78. 


93.  A  trapezoid  is  a  quadri- 
lateral which  has  only  two  of  its 
sides  parallel.  Fig.  79  is  a  trape- 
zoid. 


94.  A  trapezium  is  a  quad- 
rilateral having  no  two  sides 
parallel.  Fig.  80  is  a  trape- 
zium. 


FIG.  79. 


FIG.  80. 


§4.  GEOMETRY  AND  MENSURATION.  2!) 

95.  The  altitude  of  a  parallelogram,  or  of  a  trapczoid, 
is  the  perpendicular  distance  between  the  parallel  sides. 

96.  A  diagonal  is  a  straight  line  drawn  from  the  vertex 
of  any  angle  of  a  quadrilateral  to  the  vertex  of  the  angle 
opposite;  a  diagonal  divides  the  quadrilateral  into  two  tri- 
angles.     A  diagonal  divides  a  parallelogram  into  two  equal 
and  similar  triangles. 

97.  Ilule.  —  To  find  the  area  of  any  parallelogram,  mul- 
tiply tJie  base  by  the  altitude. 

Let  b  =  length  of  base; 

h  =  altitude; 

A  =  area. 
Then,  A  =  b  Ji. 

EXAMPLE.  —  A  lot  is  4  rods  wide  and  8  rods  long,  (a)  How  many 
square  feet  does  it  contain  ?  (/>)  What  part  of  an  acre  ? 

SOLUTION.  —  (a)  Reducing  rods  to  feet,  4  rods  =  CO  feet,  and  8  rods 
=  132  feet.  Either  side  may  be  taken  as  the  base.  Applying  the 
formula, 

A  =  bh  —  66  X  132  =  8,712  square  feet.     Ans. 

(b}  As  there  are  43,560  square  feet  in  an  acre,  8,712  square  feet 
=  8,712  -=-  43,560  =  .2  =  |  acre.  Ans. 

98.  Rule.  —  To  find  the   area   of  a  trapczoid,  multiply 
one-Jialf  tJie  sum  of  tJic  parallel  sides  by  the  altitude  of  the 
trapczoid. 

Let  a  and  b  represent  the  lengths  of   the  parallel   sides, 

and  h  the  altitude  ; 

,    la      b 
then,  A  =  h 


EXAMPLE. — How  many  square  feet  of  floor  space   are  there   in  a 
5-story  warehouse  having  the  dimensions 
given  in  Fig.  81  ? 

SOLUTION.- — As  either  side  may  be  called^ 
a,  let  it  be  the  shorter;  then,  applying  the  > 
formula,  * 

^  =  *(£^)=87.6(li^t?) 


=  27.6X51.9  =  1,432.44  square  feet.  pm 

Therefore,  for  5  stories,  there  are  1,432.44 
square  feet  X  5  =  7,162.2  square  feet  of  floor  space.     Ans. 


30 


GEOMETRY  AND  MENSURATION. 


POLYGONS. 

99.     Rule.  —  To  find  the  area  of  a  regular  polygon,  divide 
the  figiirc  into  isosceles  triangles,  compute  the  area  of  one 
triangle,   and  multiply   by   the   number   of  triangles.      The 
result  ^cvill  be  the  area  of  the  polygon. 
Let  ;/  =  number  of  sides  of  polygon  ; 
/  =  length  of  one  side  ; 

h  =  perpendicular  distance  from  the  center  to  a  side  ; 
A  =  area  of  polygon. 

n  lh 


OM- 
Then, 


1 
A  — 


2 


EXAMPLE.  —  Find  the  floor  surface  of  an  octagonal  room,  the  length 
of  whose  sides  is  5  feet.  The  distance  between  parallel  sides  is  12  feet 
1  inch  nearly. 

SOLUTION.  —  As  the  room  is  a  regular  octagon,  n  =  8,  /  =  5  feet, 
and  h  —  \  of  12  feet  1  inch  =  \  of  12.08  feet  =  6.04  feet.  Applying 
the  formula, 

nlh        8X5X6.04 


. 

A  — 


=  120.8  square  feet.     Ans. 


B 


1OO.  Rule. —  To  find  the  area  of  an  irregular  polygon, 
or  any  figure  bounded  by  straight  lines,  divide  the  figure  into 
triangles,  parallelograms,  and  trapezoids,  and  find  the  area 
of  each.  The  sum  of  these  partial  areas  will 'be  the  area  of 

the  figure. 

EXAMPLE. — A  farmer  fenced 
in  a  piece  of  land  having  the 
dimensions  given  in  Fig.  82. 
How  many  square  feet  does  it 
contain  ? 

SOLUTION. — The  distances 
marked  on  Fig.  82  being 
measured,  the  partial  areas 
are  found  to  be : 

=  325  square  feet  • 
=  720  square  feet ; 
=  840  square  feet ; 
=  237  square  feet. 


& 

79*6 


The  area  of  the  field  is,  therefore,  the   sum  of  these,  or  2,122  square 
feet.     Ans. 


•§4  GEOMETRY  AXD  MENSURATION.  31 

THE   CIRCLE. 

101.  The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  an    incommensurable   number,   and  is  usually 
denoted    in    technical    books   by  the    Greek   letter    -    (pro- 
nounced //).     The  approximate  value  of  the  ratio,  correct 
to  four    decimal  places,    is  3.1410;    hence,   approximately 
-  =  3.1410.       For   offhand    calculations,    the    ratio    is    fre- 
quently taken  as  3i  ;  that  is,  the  circumference  is  3  1  times 
the  diameter. 

102.  Rule.  —  To  find  the  circumference  of  a  circle,  mul- 
tiply the  diameter  by  3.  1416. 

To  find  the  diameter  of  a  circle,  divide  the  circumference 
by  3.1416. 

103.  Denoting  the  circumference  by  c,  the  diameter  by 
d,  and  the  radius  by  r, 

e   =  -  d  =  '2  -  r  ; 


and  r  =  —  . 

EXAMPLE.  —  A  wjieehvright  wishes  to  cut  a  length  of  tire  iron  long 
enough  to  go  around  a  4J-foot  wagon  wheel.  How  long  must  he  cut  it, 
allowing  4  inches  for  welding  ? 

SOLUTION.  —  Here  d  =  4±  feet.     Applying  the  formula, 

c  =  TT  d  -  3.  1416  X  4|  =  14.  14  feet. 

Adding  the  4  inches  =  .33  foot,  the  length  required  is  14.47  feet.  As 
this  is  only  .03  foot  =  f  inch  less  than  14.V  feet,  he  would  cut  off  14^ 
feet,  and  make  the  weld  4|  inches. 

1O4.  As  the  circle  is  divided  into  300  degrees,  and  the 
length  of  the  circumference  is  2  -  r,  the  length  of  1  degree  is 

;  or,  dividing  both  terms  by  2  -  (=  0.2832),  this  reduces 


to  —  —  ,  very  closely.     If  the  angle  is  57.3°,  the  length  of  the 
07.0 

arc  is  equal  to  the  radius;  this  angle  is  called  a  radian. 

1O5.     Rule.  —  To  find  the  length  of  an  arc  of  a  circle, 
multiply  the  number  of  degrees  in  the  are  by  the  radius,  and 
divide  by  57.296  (or,  sufficiently  close,  57.3). 
1-n 


GEOMETRY  AND  MENSURATION. 


Let  /  =  length  of  arc ; 

n  =  number  of  degrees  in  arc ; 
r  =  radius  of  arc. 

OM,  .        rxti 

Then,  /  :  =  _. 

If  the  angle  contains  degrees,  minutes,  and  seconds,  reduce 
them  to  degrees  and  decimals  of  a  degree,  thus: 

37°  30' 15"  =  37°  +  t|°  +  imhr0  =  37° +  .5° +  .004°  =  37.504°. 

EXAMPLE. — In  Fig.  83  is 
shown  a  segmental  stone 
arch  having  a  radius  of  7 
feet  1  inch  =  85  inches.  The 
angle  A  O  B  is  50°.  If  there 
are  13  ring  stones  in  the  arch, 
what  will  be  the  width,  as 
E  F,  of  each  stone  ? 

SOLUTION.  —  Here  A  D  B 
=  /;  r  =  85  inches,  and  » 
=  50°.  Applying  the  for- 

rn         85  X  50 

mula     /  =  =  — — — 

57.3  57.3 

=  74.17  inches.    Dividing  by 


the  number  of  stones, 

=  5.70  inches  = 
nearly.     Ans. 


74.17 
13 

inches, 


1O6.  When  the  chord  of  the  arc  and  the  height  of  the 
segment  (that  is,  A  B  and  CD,  Fig.  84)  are  known,  the 
length  of  the  arc  may  be  found  by  the  following  formula, 
in  which 

r  =  radius  of  arc; 

/  =  length  of  the  arc  =  length  otADB, 

Fig.  84; 

c  =  length  of  the  chord  =  length  of  AB; 
Ji  =  height  of  segment  =  length  of  CD; 


FIG.  84. 


GEOMETRY  AND  MENSURATION. 


If  the  chord  A  B  and  radius  O  A  are  given,  the  height 
CD  of  the  segment  may  be  found  by  the  formula 

//  =  r  —  Wlr'  —  c"-. 

EXAMPLE. — If  in  Fig.  83  the  span  A  B  is  6  feet,  or  72  inches,  and  the 
rise  G  D  is  8  inches,  what  is  the  length  of  the  intrados  (arc  ,•/  D  B)  of 
the  arch  ? 

SOLUTION. — Here  the  span  A  B  is  the  chord,  and  the  rise  is  the 
height  of  the  segment.  Applying  the  formula, 

72  _  4  X  73. 76  -  72 


3  3  -«-  =7135 

inches  =  74|  inches,  nearly.     Ans. 

1O7.     When  the  quotient  obtained  by  dividing  the  chord 
by  the  height  is  less  than  4.8,  that  is,  when  -j  is  less  than  4.8, 

the  formula  does  not  work  well,  the  results  not  being  suffi- 
ciently exact.     In  such  a  case,  bisect  the  c 
arc,  and  then  apply  the  formula.      Thus, 


A  B 
in  Fig.  85,  suppose  that  -^-    is  less  than 


4.8;  then  we  should  bisect  A  C  B  by 
drawing  O  C  perpendicular  to  A  B.  We 
then  find  the  length  of  the  arc  A  C  and 
multiply  the  result  by  2.  But,  in  order 
to  find  the  length  of  the  arc  A  C,  we 
must  know  the  length  of  the  radius  O  A  and  the  height  of 
the  segment  included  between  the  chord  A  C  and  the  arc 
A  C.  These  may  be  found  by  means  of  the  following  for- 
mulas, in  which 

r    —  radius  O  A  of  the  arc ; 

C  =  chord  A  B  of  the  arc ; 

c    —  chord  A  C  of  half  the  arc ; 

H  =  height  CD  of  the  segment; 

h    =  height  of  segment  included  between  chord  A  C 
and  arc  A  C. 

(a)     r  =C- 


(p)      c  = 
(0     A  = 


2  +  4 


34 


GEOMETRY  AND  MENSURATION. 


FIG. 


8^7 

Now  applying  formula 
c  =  ii/( 


EXAMPLE. — The  segmental  arch  in 
Fig.  86  has  a  span  of  60  inches  and 
a  rise  of  20  inches.  What  is  the 
length  of  its  intrados  AC B  1 

SOLUTION.— Since  60-5-20  =  3,  a 
number  smaller  than  4.8,  we  must 
find  the  length  A  C  of  half  the  arc. 
Applying  formula  (a)  to  find  the 
radius, 


=  32.5  inches. 


2  =  36.06  inches. 


8X  20 


Applying  formula  (c), 


h  =  r-£|/16r2-C2-4//2  =  32.5  -  }|/16  X  32. 52  -  GO2  -  4  X  20" 
=  5.46  inches,  nearly. 

Applying  the  formula  of  Art.  1O6, 


/  = 


+  4x5.462-36.06 


3  3 

=  38.21  inches,  length  of  arc  A  C. 

Therefore,  38.21  X  2  =  76.42  inches,  nearly  =  length  of  arc  A  C B. 

Ans. 

1O8.  Rule. —  To  find  the  area  of  a  circle,  square  the 
diameter  and  multiply  by  .7854;  or,  square  the  radius  and 
multiply  by  3. 1416. 

Let  d  =  diameter  of  circle ; 

r  =  radius  of  circle ; 
A  =  area  of  circle. 


Then, 


A  =  \*d*  -  .7854^'; 
A  —  -r*  —  3. 1416  r*. 


EXAMPLE. — What  is  the  area  of  a  circle  whose  radius  is  14  inches  ? 
SOLUTION. — Applying  the  formula, 

A  =  3.1416  X  142  =  615.75  square  inches.     Ans. 

EXAMPLE. — The  contract  for  a  brick  warehouse  provides  that  all 
openings  over  4  feet  wide  shall  be  deducted.  What  must  be  deducted 
for  a  semicircular  arch  4  ft.  8  in.  in  diameter  ? 


§4     GEOMETRY  AND  MENSURATION.      35 

SOLUTION.— Here  d  =  4  feet  8  inches  =  4g  feet.  Applying  the  for- 
mula, 

A  =  .7854^2  =  . 7854  X  (45)"  =  .78-54x21.78  =  17.11  square  feet. 
For  a  half  circle  the  area  is  -^—  =  8.55  square  feet,  practically  Si- 
square  feet.     Ans. 

1O9.  llule. —  To  find  the  diameter  of  a  circle,  t lie  area 
being  given,  divide  the  area  by  .  7854  and  extract  the  square 
root  of  t  lie  quotient. 


.  785-4 

EXAMPLE. — -To  supply  a  certain  quantity  of  water,  it  is  necessary  to 
have  a  pipe  with  an  area  of  28  square  inches.  What  will  be  the  diam- 
eter of  the  pipe  ? 

SOLUTION. — Applying  the  formula, 


/A  /  28 

=  l/:7854  =  V  .-7^54  = 


As  this  is  almost  6  inches,   a  pipe  of  that  diameter  would  be  chosen. 

Ans. 

11O.  Rule.  —  To  find  the  area  of  a  flat  circular  ring, 
subtract  the  area  of  the  smaller  circle  from  that  of  the 
larger. 

Let  d  =  the  longer  diameter; 

d^  =  the  shorter  diameter  ; 
A  =  area  of  ring. 

Then,      A  =  .7854  d*  -  7854  d*  =  .7854(^2-<3). 

EXAMPLE.—  What  is  the  sectional  area  of  a  brick  stack  whose  external 
and  internal  diameters  are  6  feet  6  inches,  and  4  feet  respectively  ? 

SOLUTION.  —  Here  d  —  6.5  feet,  and  d^  —  4  feet.  Applying  the 
formula, 

A  =  .7854(6.5*  -4")  =  .7854  X  26.25  =  20.62  square  feet.     Ans. 

If  one  diameter  and  the  area  of  the  ring  are  known,  the 
other  diameter  may  be  found  by  adding  to  or  subtract- 
ing from  the  area  of  the  given  circle  that  of  the  ring, 
and  finding  the  diameter  corresponding  to  the  resulting 
area. 


36  GEOMETRY  AND  MENSURATION.  §  4 

111.  Rule.  —  To  find  the  area  of  a  sector  of  a  circle, 
divide  the  number  of  degrees  in  the  arc  of  a  sector  by  360. 
Multiply  the  result  by  the  area  of  the  circle  of  which  the 
sector  is  a  part. 

Let  n   =  number  of  degrees  in  arc  ; 

A  =  area  of  circle  ; 
r  =  radius  of  circle  ; 
a   =  area  of  sector. 

Then,  a  =  ^4  =  .  0087267  nr\ 


EXAMPLE.  —  A  circular  window  4  feet  in  diameter  is  divided  by  radial 
ribs  into  12  equal  sections.  What  is  the  area  of  each  space  ? 

OCA° 

SOLUTION.  —  Here  n  =  —  ^-  =  30°.     Applying  the  formula, 

nA        30X.7854X42 
a  =    360   =          -MO"          =  L05  Square  feet     AnS" 

Or,  a  =  .0087267  n  r'2  =  .0087267  X  30  X  4  =  1.05  square  feet.    Ans. 

112.     Thfc  area  may  also  be  found  as  follows: 
Rule.  —  To  find  the  area  of  a  sector,  multiply  one-half  of 
the  length  of  the  arc  by  the  radius. 

Let  /  =  length  of  the  arc; 

r  =  radius  of  the  arc  ; 
a  =  area  of  sector. 

Then,  a  =  \lr. 

EXAMPLE.—  In  the  sector  O  A  D  B,  Fig.  84,  the  radius  (O  A  —  O  D) 
of  the  arc  is  6  inches,  and  the  length  of  the  chord  A  B  is  7  inches  ; 
what  is  the  area  of  the  sector  ? 

SOLUTION.  —  In  order  to  find  the  area,  we  must  first  find  the  length  of 
the  arc,  by  applying  the  formula  given  in  Art.  1O6  ;  but  before  apply- 
ing this  formula,  we  must  find  the  height  CD  of  the  segment.  Since 


AC  =  \AB  =  \X1  =  3.5,  OC  =  V~OA*-  ^C~*  =   t/63-3.53  =  4.87 
inches.     Then,  CD  —  O  D-O  C  =  6-4.87  =  1.13  inches. 
Applying  the  formula  of  Art.  1O6, 


,  .-          _  J0  . 

/  =  -  -  =  7.48  inches. 

o 

Now  applying  the  formula  given  above, 

a  =  $/r  =  $X  7.48  X  6  =  22.44  inches.     Ans. 


§4     GEOMETRY  AND  MENSURATION. 

113.  Rule. —  To  find  the  area  of  a  segment  of  a  i 
find  the  area  of  the  sector  of  which  the  segment  is  a 
and  from  this  area  subtract  the 
area  of  the  triangle  formed  by 
drawing  radii  to  the  extrem- 
ities of  the  chord  of  the  segment. 
The  result  is  the  area  of  the  seg- 
ment. 

EXAMPLE. — What  is  the  area  of  the 
upper  pane  of  glass  in  the  window 
shown  in  Fig.  87  ? 

SOLUTION. — The  radius  is  found 
by  applying  the  formula 


ircle, 
part, 


P,G  sr 


c"  +  4  //-• 


f  =  3  feet  6  inches  =  42  inches,  and  //  =  (i|    inches  =  6.70    inches. 
Substituting, 

42-  +  4x6.  75- 


8X  6-75 


—  36  inches. 


The  length  of  the  arc  is 


/  = 


4 1/42- +  4x6. 75- -42 


=  44. SI  inches. 


The  area  of  the  sector  is 

Ir       44.81X36 


a  =  -x-  — 


=  800. 5H  square  inches. 


The  altitude  of  the  triangle  is  36  inches  — 6J  inches  =  29}  inches, 

42  X  29 ' 
and  its  area  is  —       —^  =  614.25  square  inches.     The  area  of  the  seg- 

a 

ment  is,  therefore,  806.58  -  614.25  =  192.33  square  inches.  The  area  of 
the  rectangular  portion  of  the  glass  is  42  X  :>6  =  1,512  square  inches; 
adding  to  this  192.33,  there  results  1,704.33  square  inches.  Ans. 


THE   KLJ.IPSK. 

114.  An  ellipse  is  a  plane  figure  bounded  by  a  curved 
line,  to  any  point  of  which  the  sum  of  the  distances  from 
two  fixed  points  within,  called  the  foci,  is  equal  to  the  sum 
of  the  distances  from  the  foci  to  any  other  point  on  the 
curve. 


38 


GEOMETRY  AND  MENSURATION. 


In  Fig.  88  let  A  and  B  be  the  foci,  and  let  C  and  D  be  any 

two  points  on  the  perimeter. 
Then,  according  to  the  above 
definition,  A  C+CB  =  A  D 
+  D  B,  and  both  these  sums 
are  also  equal  to  the  major 
axis  E  F.  The  long  diameter 
F  E  is  called  the  major  axis; 
the  short  diameter  G  D,  the 

minor  axis.     The  foci  may  be  located  from  D  or  G  as  a 

center,  and  radius  DA  =  D  B  =  \F  E. 

115.  There  is  no  exact  method  of  finding  the  perimeter 
of  an  ellipse;  but  the  following  is  close  enough  for  most 
cases  : 

Rule.  —  Multiply  the  major  axis  by  1.82,  and  the  minor 
axis  by  1.315.  The  sum  of  the  results  will  be  the  perimeter. 

Let  */?  =  major  diameter; 

d  •=  minor  diameter"; 
C  =  perimeter,  or  circumference. 

Then,  C  —  1.82  Z>  +  1.315  d. 


116.     Rule.  —  To  find  the  exact  area  of  an  ellipse,  multi- 
ply the  product  of  its  two  diameters  by  .7854- 

A  =  \*dD  =  .7854^  A 
where  A  represents  the  area,  and  D  and  d  the  two  diameters. 

EXAMPLE.—  The  soffit  (under  surface)  of  the  semi-elliptic  stone  arch 
shown  in  Fig.  89  is  to  be 
carved  at  a  cost  of  S3.  50 
per  square  foot.  If  the 
arch  is  15  inches  wide, 
what  will  the  carving 
cost? 

SOLUTION.  —  Here  D  =  8  FIG. 

feet,  and  d  —  2  X  2  =  4  feet  ;  hence,  applying  the  formula, 

C  =  1.82  D  +  1.315  d  =  1.82X8  +  1.315X4  =  19.82  feet. 


GEOMETRY  AND  MENSURATION. 


31) 


For  one-half  the  ellipse,  the  length  is  — ^  =  9.91  feet.     The  area  is 

9.91  X  1-25  (=  15  in.)  =  12.39  square  feet.     Therefore,  the  cost  of  carv- 
ing is  12.39  X  §3.50  —  §43.36.     Ans. 

EXAMPLE. — What  is  the  area  of  the  elliptical  top  of  a  table  whose 
long  and  short  diameters  are  4  and  3  feet,  respectively  ? 

SOLUTION. — Here  D  —  4  feet   and  d  =  3  feet.      Applying   the   for- 
mula, 

A  =  .7854  D  d  =  .7854  X  4  X  3  -  9.42  square  feet.     Ans. 


AREA  OF  ANY  PLANE    FIGURE. 

117.  llule.  —  To  find  tJic  area  of  any  plane  figure,  divide 
it  into  triangles,  quadrilaterals,  circles,  or  parts  of  circles, 
and  ellipses,  find  the  area  of  each  part  separately  and  add 
the  partial  areas. 

EXAMPLE.  —  Find  the  sectional  area  of  the  half-arch  and  abutment 
shown  in  Fig.  90. 

SOLUTION.  —  Divide  the  section  into    ~f  " 
parts,  as  abi  h,  e  tg  d,  and  b  g  c.    Find 
fh  =/o-Ao  =  W-  |/16'2  -  8-  =  2.  14 
feet;  then, 

area  a  b  i  h  =  12  X  5.  14  =  61.  68  sq.  ft.  ; 
area  e  i  g  d  =  8  X  4  =  32  sq.  ft.  ; 

area  bgc  =  Sxl^U  =  19.  71  sq.  ft. 

Adding  these  partial  areas,  the  sum  is 
113.39  square  feet.  The  area  of  sector 
foe  =  .0087267  nr"1  -  .0087267  X  30  X  1<52 
=  67.02  square  feet.  The  area  of  tri- 
0  =  16  —  2.14 


FIG.  90. 


=  13.86  feet;  area  = 


"I  ^    ftfi   \/   ft 


=  55.44  square  feet.    The  area  of  seg- 


ment    feh  =  67.02-55.44  =  11.58    square    feet.       Deducting    11.58 
square  feet  from  113.39  square  feet,  the  net  area  =  101.81  square  feet. 

Ans. 


40 


GEOMETRY  AND  MENSURATION. 


AREAS  OF  IRREGULAR  FIGURES. 

118.     If  the  figure   is   so  irregular  in  shape  that  the 
ordinary  rules  cannot  be  easily  applied,  the  area  may  be 


obtained  as  follows  :  Let  Fig.  91 
represent  a  drawing  of  such  an  area. 
Suppose  it  to  be  divided  by  equi- 
distant parallel  lines  into  strips,  as 
a  b,  be,  etc.  ;  then  each  of  these 
FIG.  91.  small  areas  may  be  considered  a 

trapezoid,   and  the  whole  area  is  the  sum  of  the  partial 
areas. 

From  these  considerations  we  obtain  the  following  rule: 

119.  Rule.  —  To  find  the  area  of  any  irregular  figure, 
divide  the  figure  into  any  number  of  strips  by  equidistant 
parallel  lines.  Measure  the  length  of  the  lines;  add  together 
one-Jialf  the  lengths  of  the  end  lines  and  the  lengths  of  the 
remaining  lines,  and  multiply  this  sum  by  the  distance 
between  the  lines. 

Let  a  =  length  of  one  end  line  ; 

/    =  length  of  other  end  line  ; 
b,  c,  etc.  =  lengths  of  intermediate  lines  ; 
x    =  distance  between  parallels  ; 
A  =  area. 


Then, 


A  = 


+  etc.+j+ 


The  shorter  the  distance  x  is,  the  more  accurate  will  be  the 
result. 

EXAMPLE. — Fig.  92  is  a  map  of  a  small  island  whose  area  is  to  be 
determined.  The  parallel  lines 
are  10  feet  apart,  and  are  of 
the  lengths  marked  thereon. 
Required,  the  area  of  the 
island. 

SOLUTION. — The  extreme  left  PIG.  92. 

end    of    the    island    is    a  point;   hence,   its  length,   or  a,   is  0;  /is 
4  feet. 


5  4  GEOMETRY  AND  MENSURATION.  41 

Applying  the  formula,  A  —  (  -    - — \-  b  +  etc.  ]x 

\  *          / 

/O  +  4  \ 

=  (  — g-  +10  +  30  +  31  +  31  +  34  +  36  +  37  +  37  +  35  +  25  +  13  j  X  10 

=  321  X  10  =  3,210  sq.  ft.  Ans. 


EXAMPLES  FOU  PRACTICE 

Solve  the  following  examples : 
1.  (a)  How  many  squares  (100  sq.  ft.) 
of  shingling,  and  (b)  how  many  bundles  of 
shingles.,  250  to  the  bundle,  each  shingle 
being  6  inches  wide  and  laid  5  inches  to 
the  weather,  will  be  required  for  the  gable, 
shown  in  Fig.  93,  adding  5  per  cent,  for 
waste  ?  A  (  (a)  2\  squares. 


A 

s- 


41+  bundles. 


SUGGESTION.  —  Gable  consists  of  trape- 
zoid  and  triangle.     Find  exposed  area  of  L 
each   shingle   and   number    required    per 
square  of  14,400  square  inches.  I'm.  03. 

2.  What  is  the  cost  of  plastering  the  wall  and  ceiling  of  a  semi- 
circular alcove,  6  feet  in  diameter  and  8  feet  high,  at  35  cents  per 
square  yard  ?  Ans.  $3.4S. 


SUGGESTION. — Total   area  =  area    of    semicircle  +  (length 
circumference  X  height  of  wall). 


of    semi- 


3.     What  is  the  length  of  the  intrados  A  B  E  C  D  of  the  3-centered 

arch  shown  in  Fig.  94?        Ans.   Hi. 76  ft. 

SUGGESTION. — Find  lengths  of  arcs 
B  EC,  A  />',  and  CD  separately.  A  />' 
=  C  D. 

4.  The  space  above  the  line  A  D,  in 
Fig.  94,  is  to  be  covered  with  grille-work, 
costing  §1.50  per  square  foot.  What 
will  be  the  total  cost  ?  Ans.  §65. 

SUGGESTION. — Area  required  =  area 
of  three  sectors  less  the  area  of  triangle  G  O  F.  G  F=  GO  =  FO, 
because  angles  are  all  equal ;  then  lengths  of  three  sides  are  known, 
to  find  the  area  of  triangle. 

5.     How  many  laths  4  feet  long  and  H  inches  wide,  spaced  }  inch 
apart,  transversely,  will  be  required  for  the  walls  of  a  room  14  ft.  6  in. 


42      GEOMETRY  AND  MENSURATION.     §  4 

X  10  ft.  9  in.  X  8  ft.  high,  deducting  2  doors  36  in.  X  7  ft.  and  2  windows 
2  ft.  6  in.  X  5  ft.  6  in.  ?    Add  5$  for  waste.  Ans.  883  laths. 

SUGGESTION. — Number  of  lath  per  square  foot  =  144 -H  (width  of  1 
lath  +  1  space)  X  length  of  lath  in  inches. 

6.  A  hollow  circular  cast-iron  column,  having  an  external  diameter 
of  8  inches,  is  to  carry  a  load  of  47,120  pounds,  and  must  not  be  loaded 
more  than  4,000  pounds  per  square  inch.     What  will  be  the  thickness 
of  the  metal  ring  ?  Ans.  £  in. 

SUGGESTION. — Area  of  ring  =  total  load  -=-  load  per  square  inches. 
Area  of  smaller  circle  =  area  of  8-inch  circle  —  area  of  ring ;  from  this 
find  smaller  diameter. 

7.  If  brick  masonry  will  safely  carry  10  tons  per  square  foot,  what 
must  be  the  size  of  the  square  base  of  the  column  given  in  the  previous 
example?    (Tons  of  2,000  Ib.)  Ans.  1.54  ft.,  or  18|  in.,  square. 

SUGGESTION. — Side  of  square  =  square  root  of  area  required. 


8.  A  bar  of  iron  1  yard  long  and  1  inch 
square  weighs  10  pounds.  What  will  be  the 
weight  per  yard  of  the  I  beam  shown  in  Fig.  95  ? 
Disregard  curved  edges  and  corners.  All  work  in 
fractions. 

Ans.  41|  Ib.  per  yd. 

SUGGESTION. — Area  required  =  areas  of  3  rect- 
angles +  4  triangles.  Length  /  e  =  6  inches 
—  (j5ff  X  2).  Express  results  in  256ths  of  a  square 
inch,  and  reduce  the  sum. 


9.     What  is  the  area  of  the  cross-section  of  the  dam  shown  in  Fig. 
96  ?    K  C  =   D  A',  and  O  C 
and  O  D  are  perpendicular, 
respectively,  to    K  C    and 

D  K. 

Ans.  138.95  sq.  ft. 

SUGGESTION.— Divide  area 
into  trapezoids,  rectangles, 
etc.  Area  D  M  C  K  •=  areas 
of  equal  right  triangles 
DO  K  and  KO  C  less  area 
of  sector.  The  angle  DOC 
being  63°  26'  =  63.43°,  find 
length  of  arc  D  C,  and  area 
of  sector.  Area  D  M  CK 
=  7.55  sq.  ft. 


K 


II 


GEOMETRY  AND  MENSURATION. 


43 


10.  The  segment  A  B  C  of  the  window  shown  in  Fig.  97  is  to  be 
filled    with    ornamental   glass. 

What    is    the     cost  at   86   per 
square  foot  ?  Ans.  85.64. 

SUGGESTION. — Area  =  area 
of  sector  —  area  of  triangle. 
Find  radius  A  O  and  arc  A  B  C 
by  proper  formulas ;  OD  =  ra- 
dius —  5  inches. 

11.  (a)  What  is  the  length 
of  sash  for  an  elliptical  window 
whose   axes  are  8  feet    and  4 
feet  8  inches  ?      (b)  If  the  sash 
is  4  inches  wide,  what  is  the 
exposed  area  of  the  glass  ? 

((«)     20.70ft. 


Ans.  - 


(6)    23.03  sq.  ft. 


SUGGESTION. — The  axes  of  the  glass  are  4  inches 
than  those  of  the  sash. 


Fin.  97. 

2  =  8  inches  less 


12.  In  an  octagonal  room  having  equal  sides  the  distance  between 
the  parallel  sides  is  12  feet  and  the  distance  between  opposite  corners 
is  13  feet.  How  many  feet  B.  M.  of  1-inch  white-pine  flooring  will  be 
required  ?  Add  5£  for  waste  and  poor  stuff.  Ans.  126  ft.  B.  M. 


SUGGESTION. — Length  of  side  =  2  times  base  of  triangle. 
triangles. 


Area  =  8 


13.  Pressed  brick  are  estimated  at  7  per  square  foot  of  external 
area.     How  many  will  be  required  for  a  building  20  ft.  X  40  ft.  and  20 
feet  high;  with  2  chimneys  20  in.  X  28  in.  and  8  feet  high;  and  1  chim- 
ney 20  in.  X  28  in.  and  12  feet  high  ?     Deduct  6  window  openings,  3  ft. 
X  6  ft.  6  in. ;  1  double  window,  2  openings,  each  2  ft.  8  in.  X  9  ft.  ;  1 
front  door,  4  ft.  8  in.  X  9  ft. ;  1  rear  door,  3  ft.  X  9  ft. ;  6  window  open- 
ings, 3  ft.  X  6  ft. ;  and  1  double  window,  2  openings,  2  ft.  8  in.  X  6  ft. 

Ans.   15,750  brick. 

14.  (a)  How  many  squares  (100  square  feet)  of  slating  are  there  on  a 
house  25  ft.  X  40  ft.  with  half-pitch  roof,  supposing  the  roof  to  extend 
beyond  the  walls  1  foot  (along  the  roof)  at  eaves  and  ends?    (/>)  How 
many  slates,  14  inches  wide  and  exposed  8|  inches,  will  be  required  ? 

f  (a)     15.68  squares. 
3g  \  (b)    1,897  slates. 


Ans. 


SUGGESTION. — Rise  of  roof  =  one-half  width.  Find  exposed  area 
of  each  slate  and  number  required  per  square  of  14,400  square 
inches. 


44 


GEOMETRY  AND  MENSURATION. 


THE    MENSURATION    OF    SOLIDS. 


DEFINITIONS. 

121.  A  solid,  or  body,  has  three  dimensions:  length, 
breadth,    and   thickness.     The   sides  which  enclose  it  are 
called  the  faces,  and  their  intersections  are  called  edges. 

122.  The  entire  surface  of  a  solid  is  the  area  of  the 
whole  outside  of  the  solid,  including  the  ends. 

The  convex  surface  of  a  solid  is  the  same  as  the  entire 
surface,  except  that  the  areas  of  the  ends  are  not  included. 

123.  The  volume  of  a  solid  is  expressed  by  the  num- 
ber of  times  it  will  contain  another  volume,  called  the  unit 
of  volume.     Instead  of  the  word  volume,   the  expression 
cubical  contents  is  frequently  used. 


THE  PRISM  AND  CYLINDER. 

124.  A  prism  is  a  solid  whose  ends  are  equal  parallel 
polygons  and  whose  sides  are  parallelograms. 

Prisms  take  their  names  from  their  bases.  Thus,  a  tri- 
angular prism  is  one  whose  bases  are  triangles ;  a  pentagonal 
prism  is  one  whose  bases  are  pentagons,  etc. 


125.     A  parallelepiped  is  a  prism  whose 
bases  (ends)  are  parallelograms.     Fig.  98. 


FIG.  98. 


126.     A    cube    is   a  parallelepiped    whose 
faces  and  ends  are  squares.     Fig.  99. 


FIG.  99. 


127.  The  cube  whose  edges  are  equal  to  the  unit  of 
length  is  taken  as  the  unit  of  volume  when  finding  the 
volume  of  a  solid. 


§  4     GEOMETRY  AND  MENSURATION.      45 

Thus,  if  the  unit  of  length  is  1  inch,  the  unit  of  volume 
will  be  the  cube  whose  edges  measure  1  inch,  or  1  cubic  inch; 
and  the  number  of  cubic  inches  the  solid  contains  will  be 
its  volume.  If  the  unit  of  length  is  1  foot,  the  unit  of 
volume  will  be  1  cubic  foot,  etc.  Cubic  inch,  cubic  foot, 
and  cubic  yard  are  abbreviated  to  cu.  in.,  cu.  ft.,  and  cu.  yd., 
respectively. 

128.  A  cylinder  is  a  solid  whose  ends  arc  equal  and 
similar  curved  figures.     A  circular  cylinder  is  one 

any  section  of  which,  perpendicular  to  the  axis,  is  a 
circle.  See  Fig.  100.  Unless  otherwise  expressed, 
the  word  cylinder  always  means  a  circular  cylinder. 

129.  A  right  prism   or   right  cylinder  is  one 
whose  center  line  (axis)  is  perpendicular  to  its  base.     Fl°- 10°- 

130.  The  altitude  of  a  prism  or  cylinder  is  the  perpen- 
dicular distance  between  its  two  ends. 

131.  Rule. —  To  find  tJie  area  of  tJte  convex  surface  of 
any  right  prism,  or  rig/it  cylinder,  multiply  tltc  perimeter  of 
the  base  by  the  altitude. 

Let  /  =  perimeter  of  base ; 

h  =  altitude; 
S  =  convex  surface. 

Then,  .V  =  /  //. 

To  find  the  entire  area,  add  the  areas  of  the  two  ends  to 
the  convex  areas. 

EXAMPLE. — How  many  feet,  board  measure,  of  1-inch  sheathing  are 
needed  for  a  house  20  ft.  X  28  ft.  X  18  ft.  high,  including  gables  under 
a  half-pitch  roof,  making  no  allowance  for  openings  ? 

SOLUTION.— The  perimeter  p  of  the  house  =  20x3  +  28x2  =  0<> 
feet;  h  =  18  feet;  hence,  S  —  96  X  18  =  1,728  square  feet.  Adding  2 
gables,  20  feet  base,  and  (since  the  roof  is  half  pitch)  10  feet  high  =  200 
square  feet,  the  total  is  1,928  square  feet.  As  the  sheathing  is  1  inch 
thick  the  feet  B.  M.  is  also  1,928.  Ans. 

EXAMPLE. — A  circular  iron  chimney,  7  feet  in  diameter  and  85  feet 
high,  is  to  be  covered  outside  with  a  preservative  paint.  What  will  it 
cost  to  paint  at  20  cents  per  square  yard  ? 


46      GEOMETRY  AND  MENSURATION.     §  4 

SOLUTION. — The  perimeter  p  of  the  base  is  n  d  =  21.99  feet,  and  h 
=  85  feet.  The  surface  of  the  chimney  is  21.99  X  85  =  1,869.2  square 
feet,  or  207. 7  square  yards.  The  cost  of  painting  at  20  cents  per  square 
yard  is,  therefore,  207.7  X  §-20  =  §41.54.  Ans. 

EXAMPLE. — How  many  square  feet  of  zinc  will  be  required  to  line  a 
refrigerator  having  interior  dimensions  of  4  ft.  X  6  ft.  X  8  ft.  high  ? 

SOLUTION.— The  area  of  the  sides  =  (4  X  2  +  6  X  2)  X  8  =  160  square 
feet.  Adding  for  floor  and  ceiling  (4  X  6)  X  2  =  48  square  feet,  the 
total  is  208  square  feet.  Ans. 

EXAMPLE. — What  will  be  the  cost,  at  35  cents  per  square  yard,  of 
plastering  the  walls  and  ceiling  of  an  octagonal  room,  having  sides 
5  feet  long  and  11  feet  high,  the  distance  between  the  parallel  sides 
being  12  feet  ? 

SOLUTION. — The  area  of  the  sides  is  8(5  X  H)  =  440  square  feet,  the 

area  of  the  ceiling  is -= =  120  square  feet;  the  total  is  560  square 

JO 

feet,  or  62.22  square  yards.     At  35  cents  per  square  yard,  the  cost  is 

§21.78.     Ans. 

132.  Rule. —  To  find  the  volume  of  a  right  prism,  or 
cylinder,  multiply  the  area  of  the  base  by  the  altitude. 

Let  a  =  area  of  base ; 

h  =  altitude; 
V  =  volume. 

Then,  V  =  ah. 

If  the  given  prism  is  a  cube,  the  three  dimensions  are  all 
equal,  and  the  volume  is  equal  to  the  cube  of  one  of  the 
edges. 

v 

If  the  volume  and  area  are  given,  the  altitude  =  — .     If 

a 

the  cylinder  or  prism  is  hollow,  the  volume  is  equal  to  the 
area  of  the  ring  multiplied  by  the  altitude. 

EXAMPLE. — If  brickwork  averages  21  brick  per  cubic  foot,  how  many 
brick  are  there  in  a  pier  18  inches  square  and  6  feet  high  ? 

SOLUTION. — The  sectional  area  or  a  is  1.5x1-5  =  2.25  square  feet; 
h  =  6  feet;  hence,  V  =  a  h  =  2.25  X  6  =  13.5  cubic  feet;  at  21  brick  per 
cubic  foot,  the  number  of  bricks  in  the  pier  is  13.5  X  21  =  283.5.  Ans. 

EXAMPLE. — If  cast  iron  weighs  .26  pound  per  cubic  inch,  what  length 
of  sash  weight  will  be  necessary  to  balance  a  window  weighing 
8  pounds,  the  diameter  of  the  cylindrical  weight  being  1£  inches  ? 


§  4  GEOMETRY  AND  MENSURATION.  47 

SOLUTION. — As  there  are  2  weights,  the  weight  of  each  is  4  pounds. 
The  number  of  cubic  inches  required,  or  I",  is  4-=-.2(i  —  15.4.  The 
area  a  of  a  H-inch  circle  is  1.77  square  inches;  hence,  //,  the  length,  is 

V        154 

-  =  — ~  =  8.7  in.,  practically  8]  in.     Ans. 
a         1.  <7 


THE   PYRAMID   AM)   (OXE. 

133.  A  pyramid  is  a  solid  whose  base 
is  a  polygon  and  whose  sides  are  triangles 
uniting  at  a  common  point,  called  the  vor- 
tex. See  Fig.  101. 


1  34.  A  cone  is  a  solid  whose  base  is  a  circle 
and  whose  convex  surface  tapers  uniformly  to 
a  point  called  the  vertex.  See  Fig.  H>:i. 


FIG.  102. 


135.  A  right  pyramid  or  cone  is  one  whose  axis  is 
perpendicular  to  the  base. 

136.  The  altitude  of  a  pyramid  or  cone  is  the  perpen- 
dicular distance  from  the  vertex  to  the  base. 

137.  The  slant  height  of  a  pyramid  is  a  line  drawn 
from  the  vertex  perpendicular  to  one  of  the  sides  of  the 
base.      The  slant  height  of  a  cone  is  any  straight  line  drawn 
from  the  vertex  to  the  circumference  of  the  base. 

1 38.  Rule. —  To  find  the  convex  area  of  a  right  pyramid 
or  cone,  multiply  the  perimeter  of  the  base  by  one-half  of  the 
slant  height. 

Let  /   =  perimeter  of  base ; 

s    =  slant  height ; 
A  =  convex  area. 

_ps 


Then, 

1-12 


A  - 

A  -- 


48 


GEOMETRY  AND  MENSURATION. 


EXAMPLE. — An  octagonal  church  steeple  is  60  feet  high.  The  outer 
radius  of  the  base  (see  Fig.  103)  is  6  feet,  and  the 
length  of  one  of  the  sides  of  the  base  is  4.6  feet.  Com- 
pute the  convex  surface. 

SOLUTION. — To  find  the  slant  height,  the  length 
O  A  along  the  hips  must  first  be  found.  This  is  the 
hypotenuse  of  a  right-angled  triangle  of  which  the 
altitude  is  60  feet,  the  height  of  the  tower,  and  the 
base  is  the  radius,  6  feet.  Hence  O  A  —  4/602  +  6* 
=  60.3.  Now,  in  the  triangle  O  A  B,  O  A  =  60.3 

4  6 
feet  and  A  B  =  -^-  =  2.3  feet.    The  slant  height  is, 


Then, 


therefore,  O  B  =   |/60.32-2.32  =  60.3  feet,  nearly. 
The  perimeter  of  the  base  is  4.6  X  8  =  38k£feet.    Ap- 

.       ps       36.8x60.3 
ply  ing  the  formula,  A  —  ^-  = ^ =  1,109.5 

sq.  ft.     Ans. 

139.     Rule. —  To  find  the  volume  of  a 
pyramid  or  cone,  multiply  the  area  of  the  base 
by  one-third  of  the  altitude. 
Let  a  =  area  of  base ; 

h  =  altitude; 
V  =  volume. 

V  —  a^1 
~3~' 

EXAMPLE. — The  granite  capstone  of  a  monument 
is  a  square  pyramid  having  a  base  4  feet  square 
and  an  altitude  of  6  feet.  If  granite  weighs  170  pounds 
per  cubic  foot,  what  will  be  the  freight  charges  on  the 
piece  at  20  cents  per  100  pounds  ? 
SOLUTION. — The  area  a  of  the  base  is  4x4  =  16  square  feet,  and 

the  altitude  h  is  6  feet.     Applying  the  formula,  V  =  -—  =  — 5 —  =  32 

O  O 

cubic  feet;  at  170  pounds  per  cubic  foot,  the  weight  is  32  X  170  =  5,440 
pounds  =  54.4  hundredweight.  The  freight  is  therefore  54.4  X  §-20 
=  $10.88.  Ans. 

EXAMPLE. — If  in  the  preceding  example  the  capstone  were  conical, 
having  a  base  4  feet  in  diameter  and  a  height  of  6  feet,  how  much  «less 
would  it  weigh  than  the  pyramidal  stone  ? 

SOLUTION. — The  area  of  the  base  is  .7854^  =  12.57  square  feet. 

12  57  X  6 
Using  the  formula,  the  volume  is  —  —  =  25. 14  cubic  feet,  and  the 

O 

weight  at  170  pounds  per  cubic  foot  is  4,274  pounds,  nearly.  As  the 
pyramidal  capstone  weighs  5,440  pounds,  the  difference  in  weight  is 
5,440-4,274  =  1,166  Ib.  Ans. 


FIG.  103. 


GEOMETRY  AND  MENSURATION. 


4!) 


TIIK   KHl'STt   M    OK  A   PYRAMID   OR  C'OXK. 

140.  If  a  pyramid  or  cone  is  cut  by  a  plane  parallel  to 

the  base,  so  as  to  form  two  parts,  the 
lower  part  is  called  the  frustum  of 
the  pyramid  or  cone.  See  Figs.  104 
and  105. 

The  upper  end  of  the  frustum  of 
a  pyramid  or  cone  is  called  the  upper 
base,  and  the  lower  end  the  lower 
base.  The  altitude  of  a  frustum  is 
Flo  1(M  the  perpendicular  distance  between 
the  bases. 

141.  Ilnle. —  To  find  tJie  convex  area  of  a  frustum  of  a 
rig/if  pyramid  or  rig/it   cone,  multiply  one-half  the  sum  of 
the  perimeters  of  t lie  bases  by  t lie  slant  height  of  the  frustum. 

Let  P  =  perimeter  of  lower  base; 

p  =  perimeter  of  upper  base ; 
s   =  slant  height; 
A  =  convex  area. 


Then, 


A  - 


To  find  the  entire  area,  add  to  the  convex  area  the  areas 
of  the  two  bases. 

Ex  AMi'i.K. — A  square  mansard-tower  roof  has  the  dimensions  shown  in 
Fig.  106.  To  compute  the  slate  required  to  cover  the  tower  the  convex 
area  is  required.  What  is  this  convex  area  ? 

SOLUTION. — The  tower  has  the  form  of  a 
frustum  of  a  pyramid.  6  ft.  6  in.  =  6|  feet 
and  3  ft.  6  in.  =  3J  feet.  The  perimeter  of 
the  lower  base  is  4  X  6}  =  26  feet,  and  that 
of  the  upper  base  is  4  X  3J  =  14  feet.  In 
the  triangle  A  B  C,  A  C  =  6  feet,  and  li  C 
=  U6i  ~8>)  =  H  feet-  Hence,  the  hypote- 
nuse A  It,  which  is  the  slant  height,  is 

=  6.18   feet. 
(P 


Applying    the    formula,  A  =  - 

=  (26  +  14)  X  6.18  =  1236gq 
2 


1 


Fir..  100. 


50      GEOMETRY  AND  MENSURATION.     §4 

142.     Rule. —  To  find  the  volume  of  the  frustum  of  a 
pyramid  or  cone,  add  the  areas  of  the  upper  base,  the  lower 
base,  and  the  square  root  of  the  product  of  the  areas  of  the 
two  bases;  multiply  this  sum  by  one-third  of  the  altitude. 
Let  A  =  area  of  lower  base ; 

a  =  area  of  upper  base ; 
//  =  altitude; 
V  =  volume. 

Then,  V  =  ^  (A  +  a  +  V~A~xa). 

o 

EXAMPLE.  —A  marble  monument  is  2|  feet  square  at  the  base,  1  foot 
square  at  the  top,  and  is  16  feet  high.  If  marble  weighs  160  pounds 
per  cubic  foot,  what  is  the  weight  of  the  stone  ? 

SOLUTION. — The  area  of  the  lower  base  is  2|  X  2f  =  6.25 square  feet; 
that  of  the  upper  base  is  1  square  foot.  Applying  the  formula, 

/  1 A 

V  =  J  (A  +  a  +  \/A  Xa)  =  -=•  (6.25  +  1  +  4/6.25x1)  =  52  cubic  feet; 

o  o 

at  160  pounds  per  cubic  foot,  the  weight  is  52  X  160  =  8,320  pounds. 


THE  WEDGE. 

143.  A  "wedge  is   a   solid   having  plane   surfaces,  of 

which  the  base  is  a  parallelogram, 
the  ends  are  triangles,  and  the 
sides  are  quadrilaterals  meeting 
in  a  line  parallel  to  the  sides  of 

FIG.  iw.  the  base'     See  Fi£-  107' 

144.  Rule. —  To  find  the  volume  of  a  wedge,  multiply 
together  the  width  of  the   base,  the  perpendicular   distance 
from  the  base  to  the  edge,  and  the  sum  of  the  lengths  of  the 
three  parallel  edges;  divide  the  product  by  6. 

Let     w  =  width  of  base; 

h  —  perpendicular  distance  from  base  to  edge ; 
s   =  sum  of  lengths  of  the  three  parallel  edges ; 
V  •=•  volume. 

_M  T,       w  h  s 

Then,  V  =  — g— . 

If  the  base  is  a  rectangle,  and  the  triangular  ends  are 
parallel,  the  wedge  becomes  a  triangular  prism,  and  the 
rule  for  prisms  may  be  used. 


§4     GEOMETRY  AND  MENSURATION.      51 

EXAMPLK. — Steel  weighs  .28  pound  per  cubic  inch.     How  heavy  is  a 
steel  wedge  8  inches  long,  and  l\  in.  X  3  in.  at  the  head  ? 

SOLUTION. — Here   u1  is    \\    inches,  //   is  8  inches,  s  is  y  +  3  +  3  =  9 

1  •">  X  8  X  9 
inches.     Then  I'  =  --    -  =  18  cubic  inches.     Or,  as  this  wedge 

is  a  prism,  the  volume  =  area  of  end  X  length  of  edge;   area  of  tri- 
angle =     '  =  6  square  inches;    multiplying  by  the  length  of  the 

O 

edge,  3  inches,  the  volume  is  18  cubic  inches.     The  weight  at. 28  pound 
per  cubic  inch  is  .28  X  18  =  5.04  Ib.     Ans. 


THE   PIJISMOIDAI,   FOKMt  LA. 

145.  If  any    solid  be   sliced  in   pieces   whose  adjacent 
surfaces  are  flat,  any  piece  is  called  a  plane  section  of  the 
solid.    * 

Plane  sections  are  divided  into  three  classes:  Longi- 
tudinal sections,  cross-sections,  and  right  sections.  A 
longitudinal  section  is  any  plane  section  taken  lengthwise 
through  the  solid.  Any  other  plane  section  is  called  a 
cross-section.  If  the  surface  exposed  by  taking  a  plane 
section  of  a  solid  is  perpendicular  to  the  center  line  of  the 
solid,  the  section  is  called  a  right  section.  The  surface 
exposed  by  any  longitudinal  section  of  a  cylinder  is  a 
rectangle.  The  surface  exposed  by  a  right-  section  of  a 
cube  is  a  square;  of  a  cylinder  or  cone,  a  circle;  an  oblique 
cross-section  of  a  cylinder  is  an  ellipse. 

146.  A  prismoid  is  a  solid  having  for  its  ends  \.\\oparal/cl 
plane    surfaces   which    are  connected  by 

plane  triangular  or  quadrilateral  faces. 

Thus,  the  solid  shown  in  Fig.  108  is 
a  prismoid.  The  parallel  ends  are  the 
pentagon  ABODE  and  the  quadrilateral 
F G  HI,  and  these  ends  are  connected  by 
the  triangular  face  BCH  and  the  four 
quadrilateral  faces  A  B  H  I,  A  E  F  I, 
D  EF  G,  and  C  D  G  H.  Fm.  m 

147.  Rule. —  To  find  the  volume  of  a  prismoid,  add 
together  the  areas  of  the  two  parallel  ends,  and  four  times  the 


52      GEOMETRY  AND  MENSURATION.     §  4 

area  of  a  parallel  section  midway  between  them;  multiply  the 
stun  by  one-sixth  of  the  perpendicular  distance  between  the 
parallel  ends. 

Let  A  =  area  of  one  end ; 

a  =  area  of  other  end ; 
M  =  area  of  middle  section ; 
h    =  distance  between  ends ; 
V  =  volume. 

Then,  V  =^(A+a  +  ±  M}. 

The  area  of  the  middle  section  is  not  in  general  a  mean 
between  the  end  areas,  but  the  lengths  of  its  sides  are 
means  between  the  corresponding  lengths  on  the  ends. 
This  formula,  called  the  prismoidal  formula,  is  of  very 
extended  application  and  may  be  used  to  calculate  the 
volume  of  a  prism,  cylinder,  pyramid,  cone,  frustum  of 
pyramid  and  cone,  wedge,  sphere,  and  segments  of  spheres, 
in  addition  to  the  irregularly  shaped  bodies  to  which  it  is 
ustially  applied.  For  a  pyramid,  cone,  and  wedge,  the 
upper  area  is  0;  for  a  sphere,  the  end  areas  are  both  0;  for  a 
cylinder  and  prism,  the  areas  are  all  equal. 

EXAMPLE. — What  is  the  volume  in  cubic  yards  of  a  masonry  pier  30 
feet  high,  the  upper  and  lower  rectangular  bases  being,  respectively, 
7  ft.  X  24  ft.  and  11  ft.  X  30  ft.? 

SOLUTION. — The  area  of  the  upper  base  is  7  X  24  =  168  square  feet. 
The  area  of  the  lower  base  is  11  X  30  ==  330  square  feet.  The  dimen- 
sions of  the^iddle  section  are  the  means  of  the  corresponding  dimen- 
sions of  the  bases:  therefore,  the  width  is  —  -  =  27  feet,  and  the 

to 

7  +  11 
thickness  is  — 5 —  =  9  feet.     The  area  of  the  middle  section  is  27  X  9 

iO 

=  243  square  feet.      Applying  the  prismoidal  formula,  the  volume  is 

OA  rt  QSJO 

V  =  ^(330  +  168  +  4x243)  =  7,350  cubic    feet  =  -^^  cubic  yards 
o  &  / 

=  272|  cu.  yd.     Ans. 

148.  The  prismoidal  formula,  when  used  to  obtain  the 
volumes  of  frustums  of  pyramids  and  cones,  saves  the  labor 
of  extracting  a  square  root,  as  required  by  the  ordinary  rule. 


§4     GEOMETRY  AND  MENSURATION.      53 

EXAMIM.K. — The  upper  and  lower  bases  of  the  frustum  of  a  right  cone 
are  respectively  30  inches  and  14  inches  in  diameter,  and  the  perpen- 
dicular distance  between  them  is  30  inches.  Required,  the  volume  of 
the  frustum. 

SOLUTION. — The  diameter  of  the  middle  section  is  the  mean  of  the 
diameters  of  the  ends,  or  '  -  =  22  inches. 

/v 

Area  of  lower  base          =  ..  /  =  .7854x30-  =  700.80  sq.  in. 
Area  of  upper  base  =    a   =  .7854X  14"  =  153.94  sq.  in. 

Area  of  middle  section  =  J/  =  .7854  X  22-  =  380. 1 3  sq.  in. 

Using  the  prismoidal  formula, 

V  =  8|  (706. 86  +  153. 94  +  4x380.13)  =  14,287.92  cu.  in.  =  8.27  cu.  ft. 
b  Ans. 


TIIK   SIMIKHK. 

149.     A  sphc'iv  is  a  solid  bounded  by  a  uniformly  curved 
surface,  every  point  of  which  is  equally  dis- 
tant from  a  point  within  called  the  center. 
Fig.  10!).     The  word  ball  is  commonly  used 
instead  of  sphere. 


1 5O.  Rule. —  To  find  tlic  area  of  tJic  sur- 
face of  a  sphere,    multiply  tlie  square  of  the 
diameter  by  3.  IJflG.  FIG.  io!i. 

Let  d  =  diameter  of  sphere ; 

S  —  surface. 

Then,  5  =  *  d*  =  3.141G;/'. 

EXAMPLE. — A  ball  on  a  flagstaff  is  10  inches  in  diameter  and  is  to  be 
gilded.  Deducting  20  square  inches  for  space  covered  by  attachment 
to  pole,  how  many  books  of  gold  leaf,  each  containing  25  leaves 
3|  inches  square,  will  be  required  for  gilding  it  ? 

SOLUTION. — Applying  the  formula,  S  =  -</'2  =  3. 1416  X  10  X  10 
=  314. 16  square  inches.  Deducting  20  square  inches,  the  net  surface 
is  294.16  square  inches.  Each  gold  leaf  has  an  area  of  3jJ  in.  X  3jJ  in., 
or  11.4  square  inches,  nearly;  hence  for  294.16  square  inches,  there  will 
be  needed  294.16  -f-  11.4  =  25.8  leaves,  or  1  book  and  1  leaf.  Ans. 

151.  Ilule. —  To  find  the  volume  of  a  sphere,  multiply 
the  cube  of  the  diameter  by  .5236. 


54  GEOMETRY  AND  MENSURATION.  §  4 

Let  d  •=.  diameter; 

V  =  volume. 

Then,  V  =  ~d3  =  .5236</3. 

b 

-d*        1 
Since  -    -  =  -x*d*xd,  the  volume  of  a  sphere  is  equal 

to  one-sixth  of  the  surface  multiplied  by  the  diameter. 

EXAMPLE. — What  is  the  weight  of  a  cast-iron  ball  6  inches  in  diam- 
eter, if  the  metal  weighs  .26  pound  per  cubic  inch  ? 

SOLUTION. — Applying  the  formula, 

V  -  .5236  rt"1  =  .5236  X  216  =  113.1  cubic  inches,  nearly. 
The  weight  is,  therefore,  113.1  X  -26  =  29.4  Ib.     Ans. 

152.  The  volume  of  a  spherical  shell  is  equal  to  the 
difference  in  volume  between  two  spheres  having  the  outer 
and  inner  diameters  of  the  shell. 

1 53.  Rule.  —  To  find  the  diameter  of  a  sphere  of  known 
volume,  divide  the  volume  by  .5236  and  extract  the  cube  root 
of  the  quotient. 

d=    -3/    V 


.5236 

EXAMPLE. — A  cast-iron  ball  weighing  25  pounds  is  required  for  a 
certain  purpose.  If  cast  iron  weighs  .26  pound  per  cubic  inch,  what 
will  be  the  diameter  of  the  ball  ? 

SOLUTION. — The  volume  of  the  ball  will  be  25 -=-.26  =  96.1  cubic 
inches.  Using  the  formula, 


_7  /          v  •/     W.  L  3/-io*i    F-  r-    /»o    •          1  f  i  i     •  1 

a  =  4/   =  4/  -^^  =  ylod.5  =  5.00  inches  =  5|^  in.,  nearly. 

Ans. 


SYMMETRICAL  AND  SIMILAR  FIGURES. 
154.  An  axis  of  symmetry  is  any  line  so  drawn  that, 
if  the  part  of  the  figure  on  one  side  of 
the  line  be  folded  on  this  line,  it  will 
coincide  'exactly  with  the  other  part, 
point  for  point  and  line  for  line.  Thus, 
in  Fig.  110,  if  the  upper  semicircle  be 
folded  over  on  the  diameter  CD,  it 
will  coincide  exactly  with  the  lower 
semicircle ;  also,  if  the  part  on  the  right 

of  the  diameter  A  B  be  folded  over  on  A  B,  it  will  coincide 

exactly  with  the  part  of  the  left  of  this  line. 


§  4     GEOMETRY  AND  MENSURATION.      55 

It  is  evident -from  the  above  that  a  circle  may  have  any 
number  of  axes  of  symmetry.  In  certain  cases,  however, 
a  figure  may  be  symmetrical  with  regard  to 
only  one  axis.  Thus,  the  isosceles  triangle 
A  B  C,  Fig.  Ill,  is  symmetrical  with  regard  to 
the  axis  B D,  because  the  part  BCD  would 
coincide  with  the  part  B  A  D  if  folded  over  on 
the  line/>/?;  but  no  other  axis  of  symmetry 
could  be  drawn.  A  rectangle  has  two  axes  of 
symmetry  at  right  angles  to  each  other.  A  hexagon  has  six 
axes  of  symmetry. 

155.  Similar  figures  are  those  which  are  alike  in  form. 
As  in  the  case  of  triangles,  which  have  been  considered,  two 
figures,  to  be  similar,  must  have   their  corresponding  sides 
in  proportion,  and  the  angles  of  one  equal  to  the  correspond- 
ing angles  of   the    other.      Circles   are,   of   course,  similar 
figures. 

156.  The  areas  of  tico  similar  figures  arc  to  each  other 
as  the  squares  of  any  one  dimension.     Thus,  a  regular  octa- 
gon whose  sides  are  1  inch  long  contains  4. 828  square  inches ; 
another  with  sides  4  inches   long   contains  42  or  10  times 
4.828    square   inches  =  77.25    square    inches,    for    let    A 
=  required  area ;  then^J  :  4.828sq.  in.  =  42  :  1s,  or../  —  4.82S 
X  16  square  inches. 

This  principle  often  saves  considerable  labor  in  deter- 
mining the  areas  of  similar  figures,  as,  for  instance,  the  end 
areas  of  frustums  of  pyramids,  cones,  etc. 

EXAMPLE. — Required,  the  volume  of  the  frustum  of  a  pyramid,  the 
bases  of  which  are  equilateral  triangles,  one  with  sides  6  feet  in  length, 
the  other  with  sides  2  feet  in  length.  The  altitude  of  the  frustum  is  S  feet. 

SOLUTION. — The  area  of  the  lower  triangular  base  may  be  found 

£•       ,       />       ,       f> 

from  the  formula  A  =  \/S  (s  —  a)  (s  —  b)  (s  —  c),  where  s  =  -  -  =  9, 

** 

and  the  factors  (s  —  a),  etc.  are  each  9  —  6  =  3. 

Hence,       A  =   4/9x3x3X3  =  15.588  square  feet. 

Since  the  ends  are  similar  triangles,  their  areas  are  proportional  to 
the  squares  of  the  sides ;  or,  denoting  the  area  of  the  upper  base  by  a, 
a  :  15.588  =  2* :  6». 


56 


GEOMETRY  AND  MENSURATION. 


2* 
Hence,     a  =  15.588  X  gr,  =  15.588  X  J  =  1-732  square  feet. 

A  section  midway  between  the  bases  is  evidently,  also,  an  equilateral 

6  +  2 
triangle  with  sides      g    -    =  4  feet  in  length ;  hence,  its  area  may  be 

obtained  from  the  proportion 

M:  15. 588  =  42 : 62, 

or  M  =  15.588  X  g*  =  6.9288  square  feet. 

Now  using  the  prismoidal  formula, 
V  =  1(15.588  +  1.732  +  4  X  6.928)  =  60.04  cu.  ft.     Ans. 

157.     The  cubical  contents  (and  weights)  of  similar  solids 
are  to  each  other  as  the  cubes  of  any  one  dimension. 

EXAMPLE. — If  a  cast-iron  ball  9  inches  in  diameter  weighs  100  pounds, 
what  would  a  ball  15  inches  in  diameter  weigh  ? 

SOLUTION.—     100  :  x  -  93 :  153,  or  x  =  10°  *L?;875  =  462.96  pounds, 


the  weight  of  the  larger  ball.     Ans. 


729 


EXAMPLES  FOR  PRACTICE. 

158.      Solve  Jthe  following  examples: 

1.     What  is  the  weight  of  the  cast-iron  lintel  shown  in  Fig.  112,  esti- 
mating iron  at  .26  pound  per  cubic  inch  ?  Ans.  200.8  lb.,  nearly. 

SUGGESTION. — The  lintel  consists  of  a  rectangular  base,  a  longitudinal 
rib  whose  faces  are  trapezoids,  and  two  triangular  cross  ribs. 

tr 

ac 


FIG.  112. 


2.  (a)  Figure  the  number  of  feet  B.  M.  (board  measure)  in  the  fol- 
lowing bill  of  material,  1  foot  B.  M.  being  1  foot  square  and  1  inch 
thick;  (b)  also  the  cost  at  $16.50  per  M.  (thousand)  feet. 


23  pieces  3"  X  10" ; 
3  pieces  3"  X  10" ; 

3  pieces  3"  X  10" ; 
34  pieces  3"  X    6" ; 

4  pieces  3"  X    6" ; 
38  pieces  3"  X    4" ; 


18'  6"  long. 
15'  0"  long. 
12'  0"  long. 
14'  0"  long. 

6'  0"  long. 

5'  6"  long. 


2  pieces  3"  X  6" ; 

5  pieces  3"  X  8" ; 

56  pieces  2"  X  4" ; 

81  pieces  2"  X  4" ; 

42  pieces  2"  X  4" ; 


10'  6"  long. 
18'  6"  long. 

9'  9"  long. 

9'  3"  long. 

1'  2"  long. 


SUGGESTION. — An  easy  method  to  figure  B.  M.  is  as  follows:  Note 
that  in  the  first  item,  a  board  10  inches  wide  and  1  foot  long  contains 


GEOMETRY  AND  MENSURATION. 


57 


^  or  I  of  a  board  foot  for  each  inch  of  thickness.  Hence,  a  3"  X  10" 
plank  contains  ;}  X  3  =  2.5  feet,  B.  M.  per  foot  of  length. 

(  (a)     3,338  ft.  B.  M.,  nearly. 
S'  <  (/;)     $55.08. 

3.  What  is  the  weight  of  the  boiler  front  shown  in  Fig.  113  if  the 
metal  is  1^  inches  thick,  and  east  iron  weighs 

.26  pound  per  cubic  inch  ?  Ans.   1,638  Ib. 

SUGGESTION. — First  find  net  area  of  sur- 
face; door  openings  consist  of  rectangular  por- 
tion +  circular  segment  whose  chord  and  height 
are  known. 

4.  Figure  the  cost  of  digging  and  lining  a 
well  24  feet  deep  and  3|  feet  clear  diameter, 
the   lining   to   be   1    brick   thick,   laid   without 
mortar;  allow  18  brick  per  square  foot  of  sur- 
face of  interior  of  lining.     The  excavation  is  to 
be  5|  feet  in  diameter.     The  digging  is  to  cost 
$1.25  per  cubic  yard,  and  the  brickwork   $12 
per  thousand,  finished  and  laid.      Ans.   $81.07. 

5.  Find  the  diameterof  the  ballsof  a  10-pound 

dumbbell,  the  cylindrical  handle  of  which  is  4',  Fir..  113. 

inches  long  and  1^  inches  in  diameter.  Cast  iron  weighs  .26  pound  per 
cubic  inch.  Disregard  the  small  volumes  at  ends  of  handle,  which  are 
figured  twice.  Ans.  3.16  in.  =  3£-t-  in.,  nearly. 

SUGGESTION. — Find  contents   of   handle;    one-half  of  remainder  is 
volume  of  each  ball. 

6.  A  window  weighing  10  pounds  is  to  be  balanced  by  two  2-inch 
lead-pipe  weights.     If  the  metal  is  }  inch  thick,  and  lead  weighs  .41 

pound  per  cubic  inch,  how  long  pieces  are  required  ? 

Ans.  8.9  in.,  say  9  in. 

SUGGESTION. — Find   number   of    cubic    inches    in    each 
weight.     Length  required  =  volume  -5-  area  of  ring. 

7.  A  shop  28  ft.  X  40  ft.  and  18  feet  high,  with  gables 
8  ft.  8  in.  high,  is  to  be  built  of  brick.     The  lower  walls 
are  12  inches,  and  the  gables  8  inches  in  thickness.     Deduct 
12  windows  3  ft.  X  6  ft.  4  in.,  and  4  doors  4  ft.  X  7  ft.  all  in 
12-inch  wall.     Estimating  7  bricks  to  each  square  foot  of 
wall  4  inches  thick,  how  many  thousand  bricks  will   be 
required,  with  5  per  cent,  allowance  for  breakage,  etc.  ? 

Ans.  50,303  bricks. 
SUGGESTION. — Find  areas  and  not  volumes. 

8.  (a)  What  will  be  the  weight  of  the  square  granite 
shaft  shown  in  Fig.  114,  estimating  the  stone  at  170  pounds 
per  cubic  foot?     (b}  Allowing  4,000  pounds  per  square  foot 
on  the  soil,  what  will  be  the  size  of  the  foundation  required  ? 

j(fl)     1 68,440  Ib.  =  84. 2  T. 
{  (b)     6.5  ft.  square,  nearly. 


Ans. 


58 


GEOMETRY  AND  MENSURATION. 


Query. — What  two  solids  make  up  the  monument  ? 


FlG.  115. 


9.  What  is  the  weight  per  foot  of  the  Z  bar 
column  shown  in  Fig.  115,  figuring  steel  at  .283 
pound  per  cubic  inch  ?  Neglect  rounding  of  edges 
of  Z  bars,  but  figure  the  fillets,  as  at  a.  Check  re- 
sult by  multiplying  area  of  section  by  3.4,  which 
is  the  weight  per  foot  of  steel  for  each  square  inch 
of  section.  .  j  170.37  Ib. 

S>  (  170.57  Ib. 

SUGGESTION. — Area  of  a  fillet  =  ^  difference  be- 
tween square  of  twice  the  radius,  and  circle. 


FIG.  116. 


GEOMETRY  AND  MENSURATION. 


2ir.- 


10.  What  is  the  net  area 

of  the  4-sided  pyramidal- 
tower  roof  shown  in  Fig. 
IK),  the  openings  being 
measured  along  the  slope  ? 
Ans.  71 0.68  sci.  '"*• 

11.  Find  the  number  of 
cubie  yards  of  masonry  in 
the  foundation  walls  shown 
in  Fig.  117,  the  wall  being 
18  inches   wide  and  9  feet 
high,  and  the  footing  course 
2  ft.  (i  in.  wide  and  6  inches 
thick,   projecting   6  inches 
each  side  of  the  foundation 
wall. 

Ans.   97.90  en.  yd.,  nearly. 

SrccKSTiox. —  19j!l,r  in. 
=  !.<»•'!  ft.  ;  reduce  feet  and 
inches  to  feet  and  decimals 
by  conversion  table.  Find 
the  length  of  center  line 
around  the  walls ;  for  ex- 
ample, /;  =  18  in.  +  \  the 
thickness  of  wall ;  /  =  20 
ft.  —  thickness  of  wall;  o 
=  mean  of  inside  and  out- 
side measurements,  etc. 
The  same  center  line  ap- 
plies to  both  footing  and 
wall. 

12.  Figure  the   volume 
of  each   portion    and   also 
the  total  weight  of  the  cast- 
iron  column  shown  in  Fig. 
118,  the  metal  to  be  taken 
at  .26  pound  per  cubic  inch. 
Deduct   for    rounded    cor- 
ners, but  figure  ^/as  5  inches 
square  on  top. 

ANS. — Base,  a,  597.05  cu. 
in. ;  4  triangular  ribs,  b,  90 
cu.  in. ;  cylinder,  c,  4,106.85 
cu.  in.;  4  brackets,  d+c, 
245  cu.  in.;  4  lugs,/,  2  IS.  18 
cu.  in. ;  portion  of  cap  below 


60      GEOMETRY  AND  MENSURATION.     §  4 

k  /,  373.39  cu.  in. ;   portion  above  k  /,  67  cu.  in. ;  total,  5,692.45  cu.  in. 
Weight,  1,480  Ib. 

SUGGESTION. — Deduction  of  each  rounded  corner  =  \  of  difference 
between  a  square  of  twice  the  radius  and  a  circle  of  the  given  radius. 
Top  of  caps  above  line  k  I  =  (16  in.  square  —  12  in.  square)  X  1  in. 
thick  — 4  wedges,  16  in.  X  1  in.  at  back  and  13  in.  at  edge  and  1^  in. 
long.  See  sketch  (/>).  Area  of  top  of  d  is  strictly  more  than  5  in.  X  5 
in.  by  areas  m  o  q  and  g  n  p  [see  sketch  (a)  ]  =  difference  between 
rectangle  m  o  p  n  and  segment  of  circle  oqp.  Although  not  figured 
in  the  example,  the  student  should  calculate  it,  as  a  test  of  his  knowl- 
edge. 


ARCHITECTURAL  ENGINEERING. 


INTRODUCTION. 

1.  Architectural  engineering  is  the  study  of  the  anatomy 
of  a  building.      It  differs  from  architecture  in  that  it  does 
not  deal  with  utility  and  appearance,  but  with  strength  and 
stability.    The  architectural  engineer  designs  and  assembles 
the  skeleton  or  bony  framework  of  the  structure,  while  the 
architect  plans  the  building  to  best  accomplish  its  purpose, 
and  beautifies  it,  inside  and  out,  by  covering  with  becoming 
vesture  the  frame-like  structure  erected  by  the  engineer. 
This  structure  must  comply  with  the  conditions  demanded 
by  the  plan  and  design  of  the  building,  and  adhere  closely 
to  the  general  lines  laid  down  by  the  architect. 

A  knowledge  of  engineering  is  essential  to  architects  and 
those  engaged  in  building  operations.  Lack  of  such  knowl- 
edge on  the  part  of  architects  or  builders  has  resulted  in 
lamentable  catastrophes.  The  dangers  attendant  upon 
reckless  building  have,  in  fact,  so  thoroughly  impressed 
themselves  upon  civilized  communities  that  their  govern- 
ments, state  and  municipal,  have  prepared  the  most  strin- 
gent rules  and  ordinances  to  enforce  safe  construction. 

2.  The   purpose   of    correct    structural    design    is   not 
merely  to  secure  sufficient  strength,  for,   if  the  supply  of 
material  is  unlimited,  this  result  may  be  accomplished  by  a 
person  with  a  limited  knowledge  of  the  principles  of  engi- 
neering.    The  purpose  is  to  erect  the  strongest  possible 


2  ARCHITECTURAL  ENGINEERING.  §  5 

structure  required  by  the  existing-  conditions,  with  a  mini- 
mum amount  of  material.  The  architect  or  builder  who  is 
able  to  accomplish  this  saves  a  considerable  expenditure  of 
money,  and,  at  the  same  time,  secures  the  required  strength 
and  stability. 

3.  To  make  a  successful  design  for  a  structure,  the 
engineer  must  keep  in  mind  the  following  principles  : 

1.  Every  part  of  the  structure  must  be  strong  enough  to 
carry  the  greatest  load  to  which  it  may  ever  be  subjected. 

2.  The   material   must    be   used    in    the    most    economical 
manner  consistent  with  the  conditions  demanded  and  pre- 
scribed. 

The  first  principle  implies  a  knowledge  of  the  forces  that 
may  act  on  a  building,  such  as  the  weight  of  the  materials 
composing  it,  the  effect  of  persons  or  machinery  in  motion 
in  the  building,  and  the  effect  of  winds  and  storms  on  the 
structure,  and  also  a  knowledge  of  the  properties  of  the 
available  building  materials  and  their  ability  to  resist  these 
forces. 

The  second  principle  implies  a  knowledge  of  the  relative 
cost  of  different  building  materials  and  their  adaptability 
to  different  purposes;  the  available  commercial  sizes,  and 
the  details  of  the  processes  by  means  of  which  the 
commercial  forms  of  these  materials  are  prepared  for 
their  respective  places  in  the  building,  should  also  be 
known. 

The  rolling  mills  make,  for  example,  a  set  of  standard 
sizes  of  steel  beams  which  they  are  always  prepared  to  fur- 
nish. Now,  if  the  calculations-  show  that  a  steel  beam, 
weighing  17f  pounds  per  foot,  is  just  strong  enough  to 
carry  the  load  on  a  given  floor,  it  would  be  very  poor  engi- 
neering to  specify  a  beam '  of  that  weight,  unless  it  was 
found  that  such  a  beam  is  a  standard  commercial  size.  The 
extra  cost  per  pound  of  the  unusual  size  would  add  much 
more  to  the  cost  of  the  building  than  the  added  weight  of 
material  required  by  the  use  of  the  larger  commercial  size 
nearest  that  called  for  by  the  calculation.  The  successful 


§5  ARCHITECTURAL  ENGINEERING.  3 

design  docs  not,  then,  necessarily  imply  u  structure  in 
which  each  part  is  just  strong  enough  for  the  work  required 
of  it,  but  that,  through  a  careful  selection  of  the  available 
materials,  the  most  economical  use  of  each,  consistent  with 
the  strength  and  durability  of  the  structure,  is  made. 


THE  ELEMENTS  OF  MECHANICS. 


DEFINITIONS. 

4.  Mechanics  is  that  science  which  treats  of  the  action 
of  forces  upon  bodies,  and  the  effects  which  they  produce ; 
it  treats  of  the  laws  which  govern  the  movement  and  state 
of  rest  of  bodies,  and  shows  how  these  laws  may  be  utilized. 

5.  A  force  is  that  which  produces  or  tends  to  produce 
or  destroy  motion. 

6.  Motion  is  the  opposite  of  rest,  and  indicates  a  change 
of  position  in  relation  to  some  object. 

7.  Equilibrium -is  the  state  of  a  body  when   at  rest; 
that  is,  a  body  is  said  to  be  in  equilibrium  when,  through 
the  action  and  effect  of  opposing  systems  of  forces  upon  it, 
there  is  no  tendency  to  movement  within  it. 

8.  Statics  is  that  division   of  the   science  of   mechanics 
which  treats  of  the  relation  between  the  forces  acting  on  a 
body  at  rest  or  in  equilibrium.      Statics  includes  among 
other  things  the  action  of  forces  on  the  parts  of  a  building 
or  other  similar  structure. 

9.  Dynamics  is  that  division  of  the  science  of  mechanics 
which  treats  of  the  relation  between  the  forces  acting  on  a 
body  in  motion. 

EFFECTS    OF    A   FORCE. 

10.  The  effect  of  a  force  upon  a  body  may  be  compared 
with  another  force  when  the  three  following  conditions  are 
fulfilled  in  regard  to  both  forces: 

1-13 


4  ARCHITECTURAL  ENGINEERING.  §  5 

1.  The  point  of  application,  or  point  at  which  the  force 
acts  tipon  the  body,  must  be  knoivn. 

2.  The  direction  of  the  force,  or,  ivJiat  is  the  same  thing, 
the  straight  line  along  which  the  force  tends  to  move  the  point 
of  application,  must  be  known. 

3.  The  magnitude  of  the  force,  when  .compared  with  a 
given  standard,  must  be  known. 

In  this  section  and  the  succeeding  sections,  the  unit  of 
force  will  always  be  taken  as  one  pound,  and  the  magnitudes 
of  all  forces  will  be  expressed  in  pounds. 

11.  The  fundamental  principles  of  the  relations 
between  force  and  motion  were  first  stated  by  Sir  Isaac 
Newton.  They  are  called  "Newton's  Three  Laws  of 
Motion,"  and  are  as  follows: 

I.  All  bodies  continue  in  a  state  of  rest  or  of  uniform 
motion  in  a  straight  line,  unless  acted  upon  by  some  external 
force  that  compels  a  change. 

II.  Every  motion  or  change  of  motion  is  proportional  to 
the  acting  force,    and  takes  place   in  the  direction  of  the 
straight  line  along  which  the  force  acts. 

III.  To  every  action  there  is  always  opposed  an  equal  and 
contrary  reaction. 

From  the  first  law  of  motion,  it  is  inferred  that  a  body 
once  set  in  motion  by  any  force,  no  matter  how  small,  will 
move  forever  in  a  straight  line,  and  always  with  the  same 
velocity,  unless  acted  upon  by  some  other  force  wrhich  com- 
pels a  change. 

The  deduction  from  the  second  laiv  is  that,  if  two  or  more 
forces  act  upon  a  body,  their  final  effect  upon  that  body  will 
be  in  proportion  to  their  magnitude  and  to  the  directions  in 
which  they  act.  Thus,  if  the  wind  is  blowing  due  west, 
with  a  velocity  of  50  miles  per  hour,  and  a  ball  is  thrown 
due  north,  with  the  same  velocity,  or  50  miles  per  hour,  the 
wind  will  carry  the  ball  just  as  far  west  as  the  force  of  the 
throw  carried  it  north,  and  the  combined  effect  will  be  to 
cause  it  to  move  northwest.  The  amount  of  departure  from 


ARCHITECTURAL  ENGINEERING. 


due  north  will  be  proportional  to  the  force  of  the  wind,  and 
independent  of  the  velocity  due  to  the  force  of  the  throw. 

In  Fig".  1,  a  ball  c  is  supported  in  a  cup,  the  bottom  of 
which  is  attached  to  the  lever  o  in  such  a  manner  that  a  move- 
ment of  o  will  swing  the 
bottom  horizontally  and 
allow  the  ball  to  drop. 
Another  ball  b  rests  in  a 
horizontal  groove  that  is 
provided  with  a  slit  in 
the  bottom.  A  swing- 
ing' arm  is  actuated  by 
the  spring1  d  in  such  a 
manner  that,  w  h  e  n 
drawn  back  as  shown 
and  then  released,  it  will 
strike  the  lever  o  and  the 
ball  b  at  the  same  time. 
This  gives  /;  an  impulse 
in  a  horizontal  direction 
and  swings  o  so  as  to 
allow  c  to  fall. 

On  trying  the  experi- 
ment, it  is  found  that  b 
follows  a  path  shown  by 
the  curved  dotted  line, 
and  reaches  the  floor  at 
the  same  instant  as  c, 
which  drops  vertically. 
This  shows  that  the  force 
which  gave  the  first  ball 
its  horizontal  movement 
had  no  effect  on  the  ver- 
tical force  which  com- 
pelled both  balls  to  fall 
to  the  floor,  the  vertical  force  producing  the  same  effect 
as  if  the  horizontal  force  had  not  acted.  The  second  law 
may  also  be  stated  as  follows:  A  force  has  tlic  same  effect  in 


G  ARCHITECTURAL  ENGINEERING.  §  5 

producing  motion,  wlictJier  it  acts  upon  a  body  at  rest  or 
in  motion,  and  whether  it  acts  alone  or  with  other  forces. 

The  third  law  states  that  action  and  reaction  are  equal 
and  opposite.  A  man  cannot  lift  himself  by  his  boot  straps, 
for  the  reason  that  he  presses  downwards  with  the  same 
force  that  he  pulls  upwards ;  the  downward  reaction  equals 
the  upward  action,  and  is  opposite  to  it. 

12.  A  force  may  be  represented  by  a  line;  thus,  in 
Fig.  2,  let  A  be  the  point  of  application  of  the  force;  let  the 

length  of  the  line  A  B  represent  its  magni- 

**   tndc,   and  let  the  arrowhead  indicate   the 

FIG.  2.  direction  in  which  the  force  acts,  then  the 

line  A  B  fulfils  the  three  required  conditions  in  regard  to 

point  of  application,  direction,  and  intensity,  and  the  force 

is  fully  represented. 

THE    COMPOSITION    OF    FORCES. 

13.  Parallelogram  of  Forces. — When  two  forces  act 
upon  a  body  at  the  same  time,  but  at  different  angles,  their 
final  effect  may  be  obtained  as  follows : 

In  Fig.  3,  let  A  be  the  common  point  of  application  of  the 
two  forces,  and  let  A  B  and  A  C  represent  the  magnitude 
and  direction  of  the  forces.  The  final  effect  of  the  move- 
ment due  to  these  two  forces  will  be  the  same  whether  they 
act  singly  or  together.  Let,  for 

instance,  the  line  A  B  represent  the  ^ SO^lb. 

distance  that  the  force  A  B  would 
cause  the  body  to  move;  similarly, 
let  A  C  represent  the  distance  which 
the  force  A  C  would  cause  the  body 
to  move,  when  both  forces  were  act- 
ing separately.  The  force  A  B,  act- 
ing alone,  would  carry  the  body  to  B\ 
if  the  force  A  C  were  now  to  act  upon 
the  body,  it  would  carry  it  along  the 

line  B  D,  parallel  to  A  C,  to  a.  point  D,  at  a  distance  from  B 
equal  to  A  C.     Join  C  and  D,  then  CD  is  parallel  to  A  B, 


§  5  ARCHITECTURAL  ENGINEERING.  7 

and  A  B  D  C  is  a  parallelogram.  Draw  the  diagonal  .-//). 
According  to  the  second  la\v  of  motion,  the  body  will  stop 
at  D  whether  the  forces  act  separately  or  together,  but  if 
they  act  together,  the  path  of  the  body  will  be  along  A  D, 
the  diagonal  of  the  parallelogram.  Moreover,  the  length 
of  the  line  A  D  represents  the  magnitude  of  a  force,  which, 
acting  at  A  in  the  direction  A  D,  would  cause  the  body  to 
move  from  A  to  D\  in  other  words,  A  D,  measured  to  the 
same  scale  as  A  B  and  A  C,  represents  the  magnitude  and 
direction  of  the  combined  effect  of  the  two  forces  A  B  and  A  C ". 
The  force  represented  by  the  line  A  D  is  called  the 
resultant  of  the  forces  A  B  and  A  C.  Suppose  that  the  scale 
used  was  50  pounds  to  the  inch,  then,  if  A  B  =  50  pounds, 

and  A  C  =  62±  pounds,  the  length  of  A  B  would  be  —  =  1 

Ou 

inch,  and  the  length  of  A  C  would  be  -^-  =  1^  inches.      If 

00 

the  line  A  /?  measures  If  inches,  the  magnitude  of  the  result- 
ant, which  it  represents,  would  be  If  X  50  =  87^  pounds. 

Therefore,  a  force  of  87-i  pounds,  acting  upon  a  body  at 
A,  in  the  direction  A  D,  will  produce  the  same  result  as  the 
combined  effects  of  a  force  of  50  pounds  acting  in  the  direc- 
tion A  B  and  a  force  of  02i-  pounds  acting  in  the  direction  A  C. 

14.  The  above  method  of  finding  the  resulting  action  of 
two  forces  acting  upon  a  body  at  a  common  point,  is  correct, 
whatever  may  be  their  direction  and  magnitudes.  Hence, 
to  find  the  resultant  of  two  forces  when  their  common  point 
of  application,  their  direction,  and  magnitudes  are  known : 

llule  I. — -Through  an  assumed  point  draw  two  lines  par- 
allel with  the  direction  of  the  two  forces.  With  any  scale, 
measure  from  t  lie  point  of  intersection,  in  the  direction  of  the 
forces,  distances  corresponding  to  the  magnitudes  of  the 
respective  forces,  and  from  the  points  tints  obtained  complete 
the  parallelogram.  Draw  the  diagonal  of  the  parallelogram 
from  the  point  of  intersection  of  the  two  forces;  this  diagonal 
will  represent  the  resultant,  and  its  direction  will  be  away 
from  the  point  of  intersection  of  the  two  forces.  To  find  the 


8  ARCHITECTURAL  ENGINEERING.  §  5 

magnitude  of  the  resultant,  the  diagonal  must  be  measured 
with  the  same  scale  that  is  used  to  lay  off  the  tivo  forces. 

This  method  is  called  the  graphical  method  of  the 
parallelogram  of  forces. 

EXPERIMENTAL  PROOF. — The  principle  of  the  parallelogram 
of  forces  is  clearly  shown  in  Fig.  4.  A  B  D  C  is  a  wooden 
frame,  jointed  to  allow  motion  at  its  four  corners.  The 
length  A  B  is  equal  to  the  length  CD;  also,  A  C  is  equal  to 
B  D,  and  the  corresponding  adjacent  sides  are  in  the  ratio 


FIG.  4. 


of  2  to  3.  Cords  pass  over  the  pulleys  M  and  TV7",  carry- 
ing weights  W  and  w,  of  90  and  60  pounds,  respectively. 
The  ratio  between  the  weights  is  equal  to  the  ratio  of  the 
corresponding  adjacent  sides.  A  weight  V,  of  120  pounds, 
is  hung  from  the  corner  A. 

When  the  frame  comes  to  rest,  the  sides  A  B  and  A  C  lie 
in  the  direction  of  the  cords.  These  sides  A  B  and  A  C  are 
accurate  graphic  representations  of  the  two  forces  acting 
upon  the  point  A.  It  will  be  found  that  the  diagonal  A  D 
is  vertical,  and  twice  as  long  as  A  C;  hence,  since  A  C 


§  5  ARCHITECTURAL  ENGINEERING.  <) 

represents  a  force  of  00  pounds,  A  D  will  represent  a  force 
of  2X00,  or  1:20  pounds. 

Thus,  we  see  that  the  line  A  D  represents  the  resultant  of 
the  two  forces  A  B  and  A  C;  in  other  words,  it  represents 
the  resultant  of  the  two  weights  /['and  ic.  This  resultant 
is  equal  and  opposite  to  the  vertical  force,  which  is  due  to 
the  weight  of  / ". 

Satisfactory  results  of  this  kind  will  be  secured  when  we 
have  the  proportion, 

AB\A  C  =   W\w. 

EXAMPLK. — If  two  forces  act  upon  a  body  at  a  common  point,  both 
acting  away  from  the  body,  and  the  angle  between  them  is  80"",  what  is 
the  value  of  the  resultant,  the  magnitude  of  the  two  forces  being  (50 
pounds  and  90  pounds,  respectively. 

SOLUTION. — Draw  two  indefinite  lines  having  an  angle  of  80°  between 
them  (Fig.  5).  With  any  convenient  scale,  say  10  pounds  to  the  inch, 
measure  off  A  B  =  60  H-  10  =  6  inches,  and./  C  =  90-*- 10  =  9  inches. 

Through /?  draw  //  D  parallel  to  A  C,  and  through  C  draw  CD 
parallel  to  A  B,  intersecting  at  D.  Then  draw  A  D,  and  A  D  will  be 
the  resultant;  its  direction  is  towards  the  point  D,  as  shown  by  the 
arrow. 

Measuring  A  D,  we  find  that  its  length  is  11.7  inches.  Hence,  the 
magnitude  of  the  resultant  is  11.7  X  10  =  117  pounds.  Ans. 

~L&.  Triangle  of  Forces. — The  above  example  might 
also  have  been  solved  by  the  method  called  the  triangle  of 
forces,  which  is  as  follows: 

In  Fig.  5,  suppose  that  the  two  forces  acted  separately, 
first  from  A  to  />,  and  then 
from  B  to  D,  in  the  direction 
of  the  arrows. 

Draw  A  D ;  then  A  D  is  the 
resultant  of  the  forces  A  B 
and  A  C,  since  B  D  =  A  C\ 
but  A  D  is  a  side  of  the  tri- 
angle A  B  D.  It  will  also  be  A 
noticed  that  the  direction  of  F":- 5- 

A  D  is  opposed  lo  that  of  A  B  and  B  D\  hence,  to  find  the 
resultant  of  two  forces  acting  upon  a  body  at  a  common 
point,  by  the  method  of  triangle  of  forces : 


10  ARCHITECTURAL  ENGINEERING.  §  5 

Rule  II. —  Through  any  point,  draw  a  line  to  represent  one 
of  the  forces  in  magnitude  and  direction.  At  the  extremity 
of  this  line  draw  a  second  line  parallel  to  the  line  of  action 
of  the  other  force  and  representing  this  force  in  magnitude 
and  direction.  Join  the  extremities  of  the  two  lines  by  a 
straight  line;  then  this  joining  line  will  represent  the  result- 
ant, and  its  direction  will  be  opposite  to  that  of  the  tivo 
forces. 

When  we  speak  of  the  resultant  being  opposed  in  direction 
to  the  other  two  forces,  we  mean  that,  starting  from  the 
point  where  we  began  to  draw  the  triangle,  and  tracing  each 
line  in  succession,  the  pencil  will  have  the  same  general 
direction  around  the  triangle  as  if  passing  around  a  circle, 
from  left  to  right,  or  from  right  to  left,  but  the  closing  line 
or  resultant  must  have  an  opposite  direction;  that  is,  the  two 
arrowheads,  the  one  on  the  resultant  and  the  other  on  the  last 
side,  must  point  toivards  the  intersection  of  the  resultant  and 
the  last  side. 

16.  Resultant  of  Several  Forces. — When  three  or 
more  forces  act  upon  a  body  at  a  given  point,  their  resultant 
may  be  found  by  the  following  rule : 

Rule  III. — Find  the  resultant  of  any  tzvo  forces;  treat  this 
resultant  as  a  single  force,  and  combine  it  with  a  third  force 
to  find  a  second  resultant.  Combine  this  second  resultant 
with  a  fourth  force,  to  find  a  third  resultant,  etc.  After  all 
the  forces  have  been  thus  combined,  the  last  resultant  will 
be  the  resultant  of  all  the  forces,  both  in  magnitude  and 
direction. 

EXAMPLE. — Find  the  resultant  of  all  the  forces  acting  on  the  point 
O  in  Fig.  6,  the  length  of  the  lines  being  proportional  to  the  magnitude 
of  the  forces. 

SOLUTION. — Draw  O  E  parallel  and  equal  to  A  O,  and  ^/•'parallel 
and  equal  to  B  O ;  then  O  F  is  the  resultant  of  these  two  forces,  and  its 
direction  is  from  O  to  F,  opposed  to  O  E  and  E  F.  Consider  O  F  as, 
replacing  O  E  and  E  F,  and  draw  FG  parallel  and  equal  to  O  C;  O  G 
will  be  the  resultant  of  O  F  and  FG;  but  O  Fis  the  resultant  of  O  £ 
and  E  F;  hence,  O  G  is  the  resultant  of  O  E,  E  F,  and  FG,  and  like- 
wise of  A  0,  B  O,  and  CO.  The  line  FG,  parallel  to  CO,  could  not 


ARCHITECTURAL  ENGINEERING. 


11 


be  drawn  from  the  point  O  to  the  right  of  O  A",  for  in  that  case  it  would 
be  opposed  in  direction  to  (>/•";  but  F  G  must  have  the  same  direction 
as  O  I',  in  order  that  the  resultant  may  be  opposed  to  O  /-"and  /•"(/'. 

For  the  same  reason,  draw  G  L  parallel  and  equal  to  DO.  Join  O 
and  /.,  and(?  /,  will  be  the  res  u  It  ant  of  all  the  forces  .-/  (),  />'(>,  C<\  and 
D  O  (both  in  magnitude  and  direction)  acting  at  the  point  O.  If  /,'  O 
be  drawn  parallel  and  equal  to  O  /.,  and  having  the  same  direction,  it 
would  represent  the  effect  produced  on  the  body  by  the  combined 


action  of  the  forces  .-I  O,  />'  O,  C(),  and  /)().  In  this  solution  we 
have,  for  brevity,  spoken  of  the  forces  ./  O,  /,'(),  etc.,  and  the  result- 
ants O  F,  O  (/",  and  O  L.  It  must  be  remembered,  however,  that  these 
are  merely  lines  that  represent  the  forces  in  magnitude  and  direction. 

17.  In  the  last  figure,  it  will  be  noticed  that  OR,  R  F, 
FG,  G  L,  and  LO  are  sides  of  a  polygon  ORl:(r  L,  in 
which  O  L,  the  resultant,  is  the  closing  side,  and  that  its 
direction  is  opposed  to  that  of  all  the  other  sides.  This 
fact  is  made  use  of  in  what  is  called  the  method  of  the 
polygon  of  forces. 

To  find  the  resultant  of  several  forces  acting  upon  a  body 
at  the  same  point: ' 

Rule  IV, — Let  the  several  forces  be  represented  in  direc- 
tion and  magnitude  by  lines,  as  explained  in  Art.  12. 
Through  any  point  draw  a  line  parallel  to  one  of  tlie  forces, 
and  having  the  same  direction  and  magnitude.  At  tlic  end 
of  this  line,  draw  another  line,  parallel  to  a  second  force,  and 


12  ARCHITECTURAL  ENGINEERING.  §  5 

having  the  same  direction  and  magnitude  as  this  second  force; 
at  the  end  of  the  second  line,  draiv  a  line  parallel  and  equal 
in  magnitude  and  direction  to  a  third  force.  Thus  continue 
until  lines  have  been  drawn  equal  in  magnitude,  and  having 
the  same  directions,  respectively,  as  the  lines  representing  the 
several  forces. 

The  straight  line  joining  the  free  ends  of  the  first  and  last 
lines  ivill  be  the  closing  side  of  the  polygon  ;  mark  its  direc- 
tion opposite  to  that  of  the  other  forces  around  the  polygon, 
and  it  will  represent  in  magnitude  and  direction  the  resultant 
of  all  the  forces. 

EXAMPLE. — If  five  forces  act  upon  a  body  at  angles  of  60°,  120°,  180°, 
240°,  and  270°,  towards  the  same  point,  and  their  respective  magni- 
tudes are  60,  40,  30,  25,  and  20  pounds,  find  the  magnitude  and  direction 
of  their  resultant  by  the  method  of  polygon  of  forces.* 


FIG.  7. 


SOLUTION. — From  a  common  point  O,  Fig.  7,  draw  the  lines  of  action 
of  the  forces,  making  the  given  angles  with  a  horizontal  line  through 
O,  and  mark  them  as  acting  towards  O,  by  means  of  arrowheads,  as 
shown.  Now,  choose  some  convenient  scale,  such  that  the  whole 
figure  may  be  drawn  in  a  space  of  the  required  size  on  the  drawing. 
Choose  any  one  of  the  forces,  as  A  O,  and  draw  O  ^parallel  to  it,  and 
equal  in  length  to  30  pounds  on  the  scale.  It  must  also  act  in  the  same 
direction  as  A  O.  From  F,  draw  F  G  parallel  to  B  O,  and  make  its 
length  equal  to  40  pounds  on  the  assumed  scale.  In  a  similar  manner. 


*  All  the  angles  in  the  figure  are  measured  from  a  horizontal  line  in  a  direction 
opposite  to  the  movement  of  the  hands  of  a  watch,  from  1  °  up  to  360°  . 


§5  ARCHITECTURAL  ENGINEERING.  13 

draw  (7/7,  ///.  and  IK  parallel  to  CO,  DO,  and  EO,  and  equal  on 
the  scale  to  GO,  20,  and  :>.">  pounds,  respectively.  Join  ( '  and  A"  by  ( '  A"; 
then  O  A"  will  represent  in  magnitude  and  direction  the  resultant  of 
the  combined  action  of  the  five  forces.  The  direction  of  the  result- 
ant is  opposed  to  that  of  the  other  forces  around  the  polygon 
O  FG  H 1 A',  and  its  magnitude  is  5.")*  pounds.  Ans. 

18.  If  the  resultaht  O  A",  in  Fig.  7,  were  to  act  alone 
upon  the  body  in  the  direction  shown  by  the  arrowhead 
with  a  force  of  o5J  pounds,  it  would  produce  exactly  the 
same  effect  as  the  combined  action  of  the  five  forces. 

If  OF,  FG,  G  //,  ///,  and  IK  represent  the  distances 
and  directions  that  the  forces  would  move  the  body,  if  act- 
ing separately,  O  K  is  the  direction  and  distance  of  move- 
ment of  the  body  when  all  the  forces  act  together. 

From  what  has  been  said  before,  it  is  evident  that  any 
number  of  forces  acting-  on  a  body  at  the  same  point,  or 
having-  their  lines  of  action  pass  through  the  same  point, 
can  be  replaced  by  a  single  force  (resultant),  whose  line  of 
action  shall  pass  through  that  point. 

Heretofore,  it  has  been  assumed  that  the  forces  acted 
upon  a  single  point  on  the  surface  of  the  body,  but  it  will 
make  no  difference  where  they  act,  so  long  as  the  lines  of 
action  of  all  the  forces  intersect  at  a  single  point  either 
within  or  without  the  body,  only  so  that  the  resultant  can 
be  drawn  through  the  point  of  intersection.  If  two  forces 
act  upon  a  body  in  the  same  straight  line  and  in  the  same 
direction,  their  resultant  is  the  sum  of  the  two  forces;  but 
if  they  act  in  opposite  directions,  their  resultant  is  the 
difference  of  the  two  forces,  and  its  direction  is  the  same  as 
that  of  the  greater  force.  If  they  are  equal  and  opposite, 
the  resultant  is  zero,  or  one  force  just  balances  the  other. 

EXAMIM.K. — Find  the  resultant  of  the  forces  whose  lines  of  action 
pass  through  a  single  point,  as  shown  in  Fig.  8. 

SOLUTION. — Take  any  convenient  point  g,  and  draw  a  line  gf, 
parallel  to  one  of  the  forces,  say  the  one  marked  4^,  making  it  equal  in 
length  to  40  pounds  on  the  scale,  and  indicate  its  direction  by  the 
arrowhead.  Take  some  other  force — the  one  marked  .»?  will  be  con- 
venient; the  line  fe  represents  this  force.  From  the  point  e  draw  a 
line  parallel  to  some  other  force ;  say,  the  one  marked  JO,  and  make  it 


14 


ARCHITECTURAL  ENGINEERING. 


equal  in  magnitude  and  direction  to  that  force.  So  continue  with  the 
other  forces,  taking  care  that  the  general  direction  around  the  polygon 
is  not  changed.  The  last  force  drawn  in  the  figure  is  b  a,  representing 
the  force  marked  25.  Join  the  points  g  and  a ;  then,  g  a  represents  the 
resultant  of  all  the  forces  shown  in  the  figure.  Its  direction  is  from  g 


FIG.  8. 


to  a,  opposed  to  the  general  direction  of  the  others  around  the  polygon. 
It  does  not  matter  in  what  order  the  different  forces  are  taken,  the 
resultant  will  be  the  same  in  magnitude  and  direction,  if  the  work  is 
done  correctly. 

The  various  methods  of  finding  the  resultant  of  several 
forces  are  all  grouped  under  one  head :  the  composition  of 
forces. 


THIS    RESOLUTION    OF    FORCES. 

19.  Since  two  forces  can  be  combined  to  form  a  single 
resultant  force,  we  may  also  treat  a  single  force  as  if  it  were 
the  resultant  of  two  forces  whose  action  upon  a  body  will 
be  the  same  as  that  of  a  single  force.  Thus,  in  Fig.  9,  the 
force  O  A  may  be  resolved  into  two  forces,  O  B  and  B  A. 

If  the  force  O  A  acts  upon  a  body,  moving  or  at  rest  upon 


§5  ARCHITECTURAL  ENGINEERING.  15 

a  horizontal  plane,   and  the  resolved  force   (~>  />'  is  vertical, 

and  BA   horizontal,  OB,    measured   to   the   same   scale   as 

OA,   is    the    magnitude    of 

that     part     of     OA    which 

pushes  the  body  doi.vnwards, 

while  B  A  is  the  magnitude 

of  that  part  of  the  force  O  A 

which  is  exerted  in  pushing 

the  body  in  a  horizontal  di- 

rection.      OB  and  BA   are 

called  the  components  of  the  force    O  A,  and   when   these 

components    are  vertical  and  horizontal,   as  in  the  present 

case,  »they  are  called,   respectively,   the   vertical  component 

and  the  horizontal  component  of  the  force  O  A. 

2O.  It  frequently  happens  that  the  position,  magnitude, 
and  direction  of  a  certain  force  are  known,  and  that  it  is 
desired  to  know  the  effect  of  the  force  in  some  direction 
other  than  that  in  which  it  acts.  Thus,  in  Fig.  Ji,  suppose 
that  OA  represents,  to  some  scale,  the  magnitude,  direction, 
and  line  of  action  of  a  force  acting  upon  a  body  at  A,  and 
that  it  is  desired  to  know  \vhat  effect  O  A  produces  in  the 
direction  B  A.  Now7,  B  A,  instead  of  being  horizontal,  as 
in  the  illustration,  may  have  any  direction.  To  find  the 
value  of  the  component  of  O  A  which  acts  in  the  direction 
B  A,  we  employ  the  following  rule: 

Rule. — From  one  extremity  of  the  line  representing  tJie 
given  force  draw  a  line  parallel  to  the  direction  in  which  it  is 
desired  that  the  component  shall  act;  from  the  other  extrem- 
ity of  the  given  force,  draw  a  line  perpendicular  to  the  com- 
ponent first  drawn,  and  intersecting  it.  The  length  of  the 
component,  measured  from  the  point  of  intersection  to  the 
intersection  of  the  component  with  the  given  force,  will  give 
the  magnitude  of  the  effect  produced  by  the  given  force  in  the 
required  direction. 

Thus,  suppose  OA,  Fig.  9,  represents  a  force  acting  upon 
a  body  resting  upon  a  horizontal  plane,  and  it  is  desired  to 
know  what  vertical  pressure  O  A  produces  on  the  body. 


16  ARCHITECTURAL  ENGINEERING.  §5 

Here  the  desired  direction  is  vertical ;  hence,  from  one 
extremity,  as  O,  draw  O  B  parallel  to  the  desired  direction 
(vertical  in  this  case),  and  from  the  other  extremity  draw 
A  B  perpendicular  to  O  73,  and  intersecting  O  B  at  B.  Then 
O  B,  when  measured  to  the  same  scale  as  O  A ,  will  give  the 
magnitude  of  the  vertical  pressure  produced  by  O  A. 

ExAMi'LK. — If  a  body  weighing  200  pounds  rests  upon  an  inclined 
plane  whose  angle  of  inclination  to  the  horizontal  is  18°,  what  force 
does  it  exert  perpendicular  to  the  plane,  and  what  force  does  it  exert 
parallel  to  the  plane,  tending  to  slide  downwards  ? 

SOLUTION. — Let  A  B  C,  Fig.  10,  be  the  plane,  the  angle  A  being  18°, 
and  let  W  be  the  weight.  Draw  a  vertical 
B  nne  FD  =  200  pounds,  to  represent  the 
magnitude  of  the  weight.  Throtfgh  F 
draw  F E  parallel  to  A  B,  and  through  D 
draw  D  E  perpendicular  to  EF,  the  two 
lines  intersecting  at  E.  F  D  is  now  resolved 
into  two  components,  one  F  E  tending  to 
pull  the  weight  down  the  incline,  and  the 
other  ED  acting  as  a  perpendicular  pres- 
sure on  the  plane. 

Upon  measuring  F  E  with  the  same  scale 
by  which  the  weight  F  D  was  laid  off,  it  is  found  to  be  about  61.8 
pounds,  and  the  perpendicular  pressure  ED  on  the  plane  is  found  to 
measure  190.2  pounds.  Ans. 


EQUILIBRIUM. 

21.  When  a  body  is  at  rest,  the  forces  which  act  upon  it 
must  balance  one  another;  the  forces  are  then  said  to  be  in 
equilibrium.     The  most  important  of  the  forces  is  gravity. 

22.  A  body  is  in  stable  equilibrium  when,  if  slightly 
displaced  from  its  position  of  rest,  the  forces  acting  upon  it 
tend  to  return  it  to  that  position.      For  example,  a  cube,  a 
cone  resting  on  its  base,  a  pendulum,  etc. 

23.  A  body  acted  on  by  a  system  of  forces  is  in  unstable 
equilibrium  when  the  application  of  a  small  force  is  suffi- 
cient to  produce  motion.      A  cone  standing  upon  its  apex, 
an  egg  balanced  on  end  are  examples  of  bodies  in  unstable 
equilibrium. 


§  5  ARCHITECTURAL  ENGINEERING.  IT 

24.  A  force  acting-  on  a  body  tends  to  produce  motion 
in  two  ways : 

1.  It  tends  to  move  the  body  in  the  direction  of  the  line 
of  action  of  the  force. 

2.  If  a  point  in  the  body,  not  in  the  line  of  action  of  the  force, 
is  fixed,  the  force  tends  to  turn  the  body  around  that  point. 

25.  Conditions  of  Equilibrium. — Since  a  force  acting 
on  a  body  tends  to  produce  motion  in  two  ways,  the  follow- 
ing" conditions  must  be  fulfilled  in  order  that  a  body  be  in 
equilibrium : 

1.  The   resultant   of  all  the  forces  tending  to  move  the 
body  in  any  direction  must  be  zero. 

2.  The  resultant  of  all  the   forces  tending    to  turn  the 
body  about  any  point  must  be  zero. 


MOMENTS    OF    FORCES. 

26.  In  Fig.  11,  W  is  a  weight  which  acts  downwards  with 
a  force  of  10,000  pounds.  If  we  take  some  fixed. point  as  a, 
not  in  the  line  along-  which  the  weight  /Tacts,  and  connect 
the  point  a  with  the  line  of  action  of  W  by  a  rigid  arm,  so 
that  W  pulls  on  one  end  of  this  arm,  while  the  other  end  is 
firmly  held  at  a,  our  daily  experience  teaches  us  that  the 
pull  of  W  tends  to  turn  or  rotate  the  arm  around  the  point. 

Experience  also  teaches  that  the  tendency  to  rotation 
is  directly  proportional  to  the  magnitude  of  the  force, 
provided  that  the  arm 
remains  at  the  same 
length,  and  directly 
proportional  to  the 
length  of  the  arm,  if 


10  ft. 


the  force  remains  con- 
stant. In  general, 
therefore,  the  rotative 
effect  is  proportional  to  looooib. 

the  product  of  the  mag-  Fic-  "• 

nitude  of  the  force  multiplied  by  the  length  of  the  lever  arm. 
This  product  is  called  the  moment  of  the  force  with  respect 


18  ARCHITECTURAL  ENGINEERING.  §  5 

to  the  point  in  question.  Thus,  in  Fig.  11,  the  moment  of  the 
force  W  with  respect  to  the  point  a,  is  the  product  obtained 
by  multiplying  the  magnitude,  10,000  pounds,  by  the  per- 
pendicular distance,  10  feet,  from  the  point  a  to  the  line  of 
action  of  W. 

27.  The  point  a  which  is  assumed  as  the  center  around 
which  there  is  a  tendency  to  rotate,  is  called  the  center,  or 
origin,  of"  moments. 

28.  The   perpendicular    distance    from    the   center   of 
moments   to  the   line    along   which   the    force    act's,  is    the 
lever  arm  of  the  force,  also  called  the  leverage  of  the 
force. 

29.  Since  the  unit  of  force  is  the  pound,  and  the  ordi- 
nary unit  of  length  is  the  foot,  the  unit  of  moment  will  be 
a  derived  unit,  the  foot-pound,  and  moments  will  usually 
be  expressed  in  foot-pounds.      In  Fig.  11,  for  example,  the 
moment    of  the    force    W  with   respect  to  the  point  a  is 
10,000x10  =  100,000  foot-pounds. 

EXAMPLE. — What  is  the  moment  of  the  force  of  10,000  pounds  whose 
line  of  action  is  a  b,  Fig.  12,  the  center  of  the  moments  being  at  d  ? 


FIG.  12. 


SOLUTION. — The  perpendicular  distance  c  d  from  the  line  of  action 
of  the  force  to  the  center  of  moments  being  5  feet,  and  the  magnitude 
of  the  force  10,000  pounds,  the  moment  is  10,000  X  5  =  50,000  foot- 
pounds. Ans.  . 


§5  ARCHITECTURAL  ENGINEERING.  19 

30.  In  Fig.  lo  the  line  of  action  of  the  force  of  10,000 
pounds  passes  directly  through  the  point  d\  consequently, 

___^  iOOOO  Ib. 

a* 

FIG.  13. 

the  perpendicular  distance  from  the  line  of  action  to-the 
point  d  is  zero,  and  there  is  no  tendency  to  rotate  around 
that  point ;  therefore,  there  is  no  motion. 

31.  The  moment  of  a  force  may  be  expressed  in  inch- 
pounds,  foot-pounds,  or  foot-tons,  .depending  upon  the  unit 
of  measurement  used   to   designate  the  magnitude  of    the 
force  and  the  length  of  its  lever  arm.      For  instance,  if  the 
magnitude  of  a  force  is  measured  in  pounds,  and  the  lever 
arm    through    which    it    acts    in    inches,   the    moment   will 
be   in    inch-pounds;    again,   if    a    force    of    10     tons    acts 
through  a  lever  arm  of  20  feet,  the  moment  of  the  force  is 
10x20  =  200  foot -tons. 

EXAMPLE. — What  is  the  moment  in  inch-pounds  of  a  force  of  S,(MM) 
pounds,  if  the  length  of  the  lever  arm  is  13  feet  ? 

SOLUTION. — Since  the  moment  is  to  be  in  inch-pounds,  the  length  of 
the  lever  arm  must  be  in  inches.  13  feet  =  13  '>'  12  =  156  inches,  and 
the  moment  is  8,000  X  156  =  1,248,000  inch-pounds.  Ans. 

32.  Equilibrium    of   Moments.— In    Art.  21    it  was 
stated  that  when  a  body  is  at  rest,  all  the  forces  acting  on 
it   balance   one   another;    this   condition    is   expressed    by 
saying:  the  forces  are  in  equilibrium.     That  there  may  be 
perfect  balance  among  the  forces,  it  is  necessary  that  there 
be  not  only  no  unbalanced  force  tending  to  move  the  body 
along  some  given  line,  but   that  there   be,  also,  no  unbal- 
anced moment,  the  effect  of  which   would   turn  the  body 
about  some  point. 

In  Fig.  14  we  have  a  beam,  or  lever,  resting  on  the  sup- 
port c\  a  force  b  of  5  pounds  acts  downwards  at  the  right- 
hand  end  of  this  lever,  and  tends  to  turn  it  around  the 
point  of  support  c,  in  the  direction  traveled  by  the  hands  of 
a  clock,  that  is,  to  produce  rijjfht-lmncl  rotation.  The 
measure  of  this  tendency  is  5x10  =  50  foot-pounds. 

1-14 


20  ARCHITECTURAL  ENGINEERING.  §5 

Another  force  a  acts  downwards  on  the  left-hand  end  of 
the  lever,  its  tendency  being  to  produce  left-hand  rotation, 
or  to  turn  the  lever  in  the  direction  opposite  to  that  traveled 
by  the  hands  of  a  clock.  Since  the  force  a  is  10  pounds,  and 
it  acts  with  a  lever  arm  of  5  feet,  its  moment  is  10  X  5  =  50 
foot-pounds,  the  same  as  the  moment  of  the  force  b.  We 
thus  have  two  equal  moments,  one  tending  to  turn  the  lever 
to  the  right,  and  the  other  to  the  left ;  as  a  result,  the  effect 

-5ft. -t- lO-ft. 


PIG.  14. 

of  one  is  neutralized  by  the  effect  of  the  other,  and  the 
second  condition  of  equilibrium  is  fulfilled.  This  con- 
dition is  expressed  by  saying:  there  is  equilibrium  of 
moments. 

33.  Positive  and  Negative  Moments. — We  may  dis- 
tinguish between  the  directions  in  which  there  is  a  tendency 
to  produce  rotation  by  the  use  of  the  signs  -(-  and  — .    Thus, 
if  a  force  tends  to  produce  right-hand  rotation,  its  moment 
may  be  called  positive  and  have  the  plus  sign,  while  a  force 
that  tends  to  produce  rotation  in  the  opposite  direction  is 
called  negative,  and  its  moment  have  the  minus  sign.    That 
there  may  be  equilibrium  of  moments,  the  above  considera- 
tions show  that  the  difference  between  the  sum  of  the  posi- 
tive moments  and  the  sum  of  the  negative  moments  must 
be  zero;  this  difference  is  called  the  algebraic  sum  of  the 
moments.     We  therefore  have  the  principle:  In  order  that 
there  may  be  equilibrium,  the  algebraic  sum  of  the  moments 
of  all  the  forces  acting  on  a  body  must  be  zero. 

34.  Resultant  Moments. — In  Fig.  15  is  shown  a  lever 
composed  of  two  arms  at  right  angles  to  each  other,  and  free 
to  turn  about  the  center  c.    A  force  a  acts  on  the  horizontal 
arm  in  such  a  manner  that  it  tends  to  produce  left-hand 


ARCHITECTURAL  ENGINEERING. 


11 


rotation,  its  moment  being  K>X-">  =  •">"  foot-pounds,  which, 
since  it  tends  to  produce  left-hand  rotation,  shall  be  called 
negative.  Another  force  />,  whose  moment  with  respect  to 
the  center  c  is  l^x:>  =  -W  foot-pounds,  tends  to  produce 
right-hand  rotation.  Considering  the  effect  of  these  two 
forces  only,  we  see  that,  to  secure  equilibrium,  there  must 
be  another  force  acting  in  such  a  manner  as  to  overcome  the 
difference  between  the  moments  of  the  two ;  that  is,  it  must 

b 12  to  L  • 


5ft 


tend  to  produce  right-hand  rotation  with  a  moment  equal  to 
—  50-J-36  =  —  14  foot-pounds.  This  moment  is  called  the 
resultant  moment  of  the  two  forces  a  and  b. 

If  the  length  of  the  lever  arm  of  the  force  which  acts  to 
produce  the  resultant  moment  is  known,  the  magnitude  of 
the  force  may  readily  be  found.  Thus,  in  the  present  case, 
the  resultant  moment  is  —  14  foot-pounds;  let  it  be  required 
to  find  the  force  to  produce  equilibrium,  when  acting  with 
a  lever  arm-  7  feet  long.  Since  the  moment  is  the  product 
of  the  force  multiplied  by  its  lever  arm,  it  follows  that  the 
required  force  may  be  found  by  dividing  the  given  moment 
by  the  length  of  the  lever  arm ;  consequently,  the  required 
force  is  14  -j-  7  =  2  pounds. 


22  ARCHITECTURAL  ENGINEERING.  §  5 

If,  instead  of  the  two  forces  just  considered,  we  have  a 
body  acted  on  by  any  number  of  forces  whose  moments 
about  a  given  center  are  known,  the  resultant  moment  of 
these  forces,  that  is,  the  moment  of  the  force  required  to 
produce  equilibrium,  is  the  algebraic  sum  of  the  moments 
of  the  given  forces ;  and,  further,  if  the  length  of  the  lever 
arm  of  the  resultant  moment  is  known,  the  magnitude  of 
the  required  force  can  be  found  by  dividing  the  moment  by 
the  length  of  the  lever  arm. 

35.  The  above  principles  may  be  expressed  as  fol- 
lows : 

Rule. —  To  find  the  force  required  to  produce  equilibrium 
of  moments,  when  the  moments  of  any  number  of  given  forces 
and  the  lever  arm  of  the  required  force  are  given,  divide  the 
algebraic  sum  of  the  given  moments  by  the  length  of  the  given 
lever  arm.  If  t  lie  algebraic  sum  is  positive,  the  tendency  of 
t  lie  required  force  is  to  produce  left-hand  rotation  ;  if  nega- 
tive, the  tendency  of  the  force  is  to  produce  right-hand 
rotation. 

EXAMPLE. — In  Fig.  16  we  have  a  system  of  forces,  shown  by  the 
arrows,  acting  in  various  directions  and  at  various  distances  from  the 
center  O.  The  force  F'  is  25  pounds  and  its  lever  arm  O  p'  is  8  feet, 
F"  is  16  pounds  with  a  lever  arm  Op"  of  12  feet,  F'"  is 40  pounds  with 

a  lever  arm  Op'"  of  6  feet,  and  the 
force  Fv  is  100  pounds,  acting  directly 
through  the  center  O.  If  the  distance 
Op"'  is  12  feet,  what  must  be  the  magni- 
tude of  the  force  Flv  in  order  to  pro- 
duce equilibrium  of  moments  ? 

SOLUTION. — As  shown  by  the  arrows, 
the  forces  tending  to  produce  right-hand 
rotation  are  F'  and  F'",  and  their 
moments,  called  positive,  are,  respectively,  25  X  8  =  +  200  foot-pounds, 
and  40  X  6  =  +  240  foot-pounds.  The  lever  arm  of  the  force  Fv  is  zero ; 
consequently,  it  has  no  moment  with  respect  to  the  center  O.  The 
force  F"  tends  to  produce  left-hand  rotation,  and  its  moment  is  16  X  12 
=  — 192  foot-pounds.  The  algebraic  sum  of  the  moments  of  the 
given  forces  is  +200  +  240  —  192  =  +248  foot-pounds;  therefore, 
according  to  the  rule,  the  force  Fiv  must  be  248  -*- 12  =  20£  pounds, 


Fro.  16. 


ARCHITECTURAL  ENGINEERING. 


which,  since  the  algebraic  sum  of  the  given  moments  is  positive,  must 
tend  to  produce  left-hand  rotation,  as  shown  by  the  arrow.     Ans. 

#(>.  The  principles  in- 
volved in  the  theory  of 
moments  are  among  the 
most  simple  in  mechanics, 
and  at  the  same  time  of 
the  greatest  practical  im- 
portance in  the  solution  of 
problems  relating  to  the 
strength  of  beams,  girders, 
and  trusses. 

EXAMPLE. — In  Fig.  17,  the 
lower  tie  member  in  the  roof 
truss  has  been  raised  to  get  a 
vaulted -ceiling  effect  in  the 
upper  story  of  the  building, 
which  the  truss  covers.  The 
weight  transmitted  through 
this  member  to  the  pier  wall  is 
30,000  pounds ;  there  is,  conse- 
quently, an  equal  upward  force 
due  to  the  reaction  of  the  wall. 
This  force  of  30,000  pounds 
tends  to  break  the  truss  by 
producing  rotation  about  the 
point  b.  What  is  its  moment 
around  the  point  b  ? 

SOLUTION. — Since  the  per- 
pendicular distance  from  the 
line  of  action  of  the  force  is  3  feet,  the  moment  of  the  force  a  around 
the  point  b  is  30,000  X  3  =  90,000  foot-pounds.  Ans. 


THE    I,EVKK. 

37.  A  lever  is  a  bar  capable  of  being  turned  about  a 
pin,  pivot,  or  point,  as  in  Figs.  18,  11»,  and  20. 

The  object  W  to  be  lifted  is  called  the  -weight ;  the  force 
Pused  is  called  the  power;  and  the  point  or  pivot  /'  is 
called  the  fulcrum. 

That  part  of  the  lever  between  the  weight  and  the  ful- 
crum, or  Fb,  is  called  the  weltfhX  arm,  and  the  part 


ARCHITECTURAL  ENGINEERING. 


between  the  power  and  the  fulcrum,  or  Fc,  is  called  the 
power  arm. 

Take  the  fulcrum,  or  point  F,  as  the  center  of  moments ; 
then,  in  order  that  the  lever  shall  be  in  equilibrium,  the 
moment  of  P  about  F,  or  PxFc,  must  be  equal  to  the 

fP 


FIG.  18.  FIG.  19. 

moment  of  W  about  F,  or  WxFb.  That  is,  PxFc 
=  WxFb,  or,  in  other  words,  the  product  of  the  power  and 
the  power  arm  is  equal  to  the  product  of  the  weight  and  the 
weight  arm. 

If  F~be  taken  as  a  center,  and  arcs  be  described  through 
b  and  c,  it  will  be  seen  that,  if  the  weight  arm  is  moved 
through  a  certain  angle,  the  power  arm  will  move  through 
the  same  angle ;  also,  that  the  distance  that  W  moves  will 
be  proportional  to  the  distance  that  P  moves.  From  this  it 
is  seen  that  the  power  arm  is  proportional  to  the  distance 
through  which  the  power  moves,  and  the  weight  arm  is 
proportional  to  the  distance  through 
which  the  weight  moves. 

Hence,   instead  of  writing  PxFc 
=   WxFb,  we  might  have  written  it 
P  X  distance  through  which  P  moves 
=  Wx  distance 
moves. 


F 


FIG.  20. 


W 


through     which    W 

This  is  the  general  law  of  all  machines,  and  can  be 
applied  to  any  mechanism,  from  the  simple  lever  up  to  the 
most  complicated  arrangement.  Stated  in  the  form  of  a 
rule  it  is  as  follows : 

Rule. —  The  power  multiplied  by  the  distance  through 
iv hie h  it  moves  is  equal  to  the  weigJit  multiplied  by  the 
distance  through  which  it  moves. 

EXAMPLE. — If  the  weight  arm  of  a  lever  is  6  inches  long  and  the 
power  arm  is  4  feet  long,  how  great  a  weight  can  be  raised  by  a  force 
of  20  pounds  at  the  end  of  the  power  arm  ? 


ARCHITECTURAL  ENGINEERING. 


SOLUTION.—  4  feet  =  48  inches.  Hence,  20x48  =  ll'x  6,  or  W 
=  160  pounds.  Ans. 

EXAMPLK. — (<?)  What  is  the  ratio  between  the  power  and  the  weight 
in  the  last  example  ?  (6)  In  the  last  example,  if  P  moves  24  inches,  how 
far  does  II'  move  ?  (c)  What  is  the  ratio  between  the  two  distances  ? 

SOLUTION. — (a)  20  :  160  =  1:8;  that  is,  the  weight  moved  is  8  times 
the  power.  Ans. 

480 

(b)  20  X  24  =  160  X  -f.     .r  =  —  =  3   inches,    the   distance   that  If 

moves.     Ans. 

(c)  3  :  24  =  1  :  8,  or  the  ratio  is  1  :  8.     Ans. 

The  law  which  governs   the  straight  lever  also  governs 

the  bent  lever;  but  care  must  be  taken  to  determine  the 

c 


\W\ 
FIG.  21.  FIG.  22. 

true  lengths  of  the  lever  arms,  which  are  in  every  case  the 
perpendicular  distances  from  the  fulcrum  to  t 'he  line  of direc- 
tion of  the  weight  or  power. 

c 


FIG.  23. 


Thus,  in  Figs.  21,  22,  23,  and  24,  Fc  in  each  case  repre- 
sents the  power  arm,  and  Fb  the  weight  arm. 


EXAMPLES  FOR  PRACTICE. 

1.  A  lever  arm  has  a  length  of  10  feet;  the  load  acting  upon  the 
end  of  the  lever  is  6,000  pounds.  What  is  the  moment  of  this  load  in 
inch-pounds  ?  Ans.  720,000. 


26  ARCHITECTURAL  ENGINEERING.  §  5 

2.  A  piece  of  timber  20  feet  long  is  balanced  at  a  point  8  feet  from 
one  end,  the  load  at  this  end  being  9,000  pounds.     What  is  the  load  at 
the  other  end  ?  Ans.  6,000  Ib. 

3.  The  one  support  of  a  beam  20  feet  long  is  8  feet  from  the  left- 
hand  end ;  at  this  end  is  a  load  of  25  pounds ;  at  the  right  of  the  sup- 
port, 3  feet  distant,  is  a  load  of  5  pounds ;  and  at  7  feet  to  the  right  is  a 
load  of  10  pounds.     What  load,  and  at  which  end  should  it  be  placed 
to  produce  balance,  or  equilibrium,  in  the  beam  ? 

Ans.  9.58  Ib.  at  right-hand  end. 

4.  A  steel  I  beam,  which  extends  6  feet  outside  of  the  center  of  a 
building  wall,  and  3  feet  inside,  is  required  to  support  a  load  upon  the 
outside  end  of  4,000  pounds.     What  load  on  the  inner  end  will  keep 
the  beam  from  tilting  ?  Ans.  8,000  Ib. 

CENTER    OF    GRAVITY. 

38.  The  center  of  gravity  of  a  body,  or  of  a  system 
of  bodies,  is  that  point  from  which,  if  the  body  or  system 
were  suspended,  it  would  be  in  equilibrium.      If  the  body 
or  system  were  suspended  from  any  other  point  than  the 
center  of  gravity,  and  in  such  a  manner  as  to  be  free  to 
turn  about  the  point  of  suspension,  the  body  would  rotate 
until  the  center  of  gravity  reached  a  position  directly  under 
the  point  of  suspension. 

39.  Center  of  Gravity  of  Plane  Figures. — If  a  plane 
figure  has  an  axis  of  symmetry,  this  axis  passes  through  its 
center  of  gravity.      If  the  figure  has  two  axes  of  symmetry, 
its  center  of  gravity  is  at  their  point  of  intersection. 

The  center  of  gravity  of  a  triangle  lies  on  a  line  drawn 
from  a  vertex  to  the  middle  point  of  the  opposite  side,  and 
at  a  distance  from  that  side  equal  to  £  of  the  length  of  the 
line.  Or,  draw  a  line  from  another  vertex  to  the  middle 
point  of  the  side  opposite,  and  the  intersection  of  the  two 
lines  will  be  the  center  of  gravity. 

The  perpendicular  distance  of  the  center  of  gravity  of  a 
triangle  from  the  base  is  equal  to  ^  of  the  altitude. 

The  center  of  gravity  of  a.  parallelogram  is  at  the  intersec- 
tion of  its  two  diagonals. 

The  center  of  gravity  of  an  irregular  four-sided  figure 
may  be  found  by  first  dividing  it  by  a  diagonal  into  two  tri- 
angles and  joining  their  centers  of  gravity  by  a  straight  line ; 


§  5  ARCHITECTURAL  ENGINEERING.  27 

then,  by  means  of  the  other  diagonal,  divide  it  into  two 
other  triangles,  and  join  their  centers  of  gravity  by  another 
straight  line;  the  center  of  gravity  of  the  figure  is  at  the 
intersection  of  the  lines  joining  the  centers  of  gravity  of  the 
two  sets  of  triangles. 

The  distance  of  the  center  of  gravity  of  the  surface  of  a  half 
circle  from  the  center  is  equal  to  the  product  of  the  radius 
multiplied  by  .42-4. 

LOADS  CARRIED  BY  STRUCTURES. 

DEAD    LOAD. 

WEIGHT  OF  lil'ILJHNG   MATKHIAL*. 

4O.  The  materials  used  in  building  construction  have 
considerable  weight.  Every  piece  of  iron,  timber,  masonry, 
and  brickwork  has  a  tendency  to  move  towards  the  center  of 
the  earth,  and  this  tendency  of  the  parts  of  the  structure  to 
fall  to  the  ground  must  be  resisted  by  the  strength  of  the 
various  members  composing  it.  The  portion  of  this  force  to 
be  resisted  by  any  member  of  the  structure  is  called  the  dead 
load ;  it  is  the  sum  of  the  weights  of  every  piece  of  material 
in  the  building  which  must  be  supported  by  that  member. 

Before  the  dead  load  can  be  computed,  the  weight  of  vari- 
ous building  materials  must  be  known.  The  following 
tables  give  the  weights  of  building  materials  in  common  use. 
The  units  in  which  these  weights  are  expressed  are  those 
most  often  used  to  make  estimates  of  loads  in  engineering 
calculations.  Thus,  Table  1  gives  the  weight  per  cubic  foot 
of  the  materials  usually  measured  by  that  unit,  together 
with  the  weight  per  cubic  inch  of  a  few  often  measured  in 
inches;  while  Table  2  gives  the  weights  of  such  materials  as 
are  used  in  the  construction  of  floors,  roofs,  ceilings,  etc., 
where  the  quantities  are  generally  expressed  in  square  feet. 
While  it  is  not  necessary  for  the  student  to  memorize  all  of 
Table  1,  it  is  well  for  him  to  keep  in  mind  the  weights  of 
the  materials  printed  in  Italic. 


ARCHITECTURAL  ENGINEERING. 


TAUTjK   1. 
WEIGHT  OF  BUILLUNG  MATERIALS. 


Name  of  Material. 


PerCu.In.     Per  Cu.  Ft. 


Average  Weight  in 
Pounds. 


Aluminum 096 

Asphalt  pavement  composition 

Bluestone 

Brass 302 

Brickwork,  in  lime  mortar 

Brickwork^  in  cement  mortar 

Bronze 319 

Cement,  Portland. 

Cement,  Rosendale 

Concrete,  in  cement 

Copper,  cast 319 

Earth,  dry  and  loose 

Earth,  dry  and  moderately  rammed.  . 

Gneiss,  common 

Gneiss,  in  loose  piles 

Gravel 

Iron,  cast 26 

Iron,  wrought 277 

Lead,  commercial  cast 412 

Limestone 

Marble 

Masonry,  granite  or  limestone 

Masonry,  granite  or  limestone  rubble . 
Masonry,  granite  or  limestone  dry  rubble 

Masonry,  sandstone 

Mortar,  hardened 

Quartz,  common  pure 

Sand,  pure  quartz,  dry 

Sandstone,  building,  dry 

Slate 

Snow,  fresh  fallen 

Steel,  structural 283 

Terra  cotta 

Terra-cotta  masonry  work 

Tile.. 


166 

130 

160 

523 

120 

130 

552 

80  to  100 
56  to  60 

140 

550 

72  to  80 
90  to  100 

168 

96 

117  to  125 

450 

480 

712 

170 

164 

165 

150 

138 

145 
90  to  100 

165 

90  to  106 

144  to  151 

160  to  180 

5  to  12 

490 

110 

112 
110  to  120 


ARCHITECTURAL  ENGINEERING. 


TAULK   2. 
AVKKiHT   OF   1H  II.IMN<;    M  ATKKI A  LS. 


Name  of  Material. 


Average 
Weight  per 
Square  Foot 
in  Pounds. 


Corrugated  galvanized  iron  No.  ^0,  unboarded . 

Copper,  1C  oz. ,  standing  seam 1-| 

Felt  and  asphalt,  without  sheathing •  2 

Glass,  ^  inch  thiek :  1 1 

Hemlock  sheathing,  1  inch  thick :>4- 

Lead,  about  £  of  an  inch  thick (j  to  <S 

Lath  and  plaster  ceiling  (ordinary) <!  to  8 

Mackite,  1  inch  thick,  with  plaster '  10 

Neponset  roofing  felt,  'Z  layers 4- 

Spruce  sheathing,  1  inch  thick 2 

Slate,  f\  inch  thick,  3  inches  double  lap 0-J 

Slate,  £  inch  thick,  3  inches  double  lap 44; 

Shingles,  6"  X  18",  $  to  weather '  •> 

Sky  light  of  glass,  T\  inch  to  ^  inch,  including  frame  4  to  10 

Slag  roof,  4-ply '  4 

Tin,  IX | 

Tiles,  10A/XGi"xjT;  5i"  to  weather  (plain) I  18 

Tiles,  14£"XKH";  7i"  to  weather  (Spanish) '  8.V 

White-pine  sheathing,  1  inch  thick 

Yellow-pine  vSheathing,  1  inch  thick 4 


41.  In  obtaining  the  dead  load  upon  roof  trusses,  it  is 
necessary,  after  having  found  the  weight  of  the  sheathing 
and  roofing,  to  add  a  certain  weight  per  square  foot,  to  rep- 
resent the  \veight  of  the  truss  or  members  supporting  the 
roof.  Not  knowing,  as  yet,  the  size  and  weight  of  the  dif- 
ferent members  in  the  roof  truss,  we  must  assume  approxi- 
mate weights. 

Table  3  gives  the  approximate  weights  of  the  trusses,  or 
principals,  as  they  are  called,  for  roofs  of  different  spans. 


30 


ARCHITECTURAL  ENGINEERING. 


These  weights  are,  of  course,  only  assumed,  and  may  not  be 
within  25  per  cent,  of  the  actual  weight  of  the  principals. 
They  are,  however,  generally  on  the  side  of  safety. 


TABLE    3. 

POUNDS  TO  BE  ADDED  FOR  THE  WEIGHT  OF  THE 
PRINCIPALS,  OR  ROOF  TRUSSES. 

Spans  up  to  40  feet,  4  pounds  per  sq.  ft.  of  area  covered. 
Spans  40  to  60  feet,  5  pounds  per  sq.  ft.  of  area  covered. 
Spans  60  to  80  feet,  6  pounds  per  sq.  ft.  of  area  covered. 
vSpans  80  to  100  feet,  7  pounds  per  sq.  ft.  of  area  covered. 

It  is  required,  in  the  application  of  Table  3,  to  obtain  the 
weight  in  pounds  per  square  foot  of  roof  surface.  As  the 
weights  given  in  the  table  are  in  pounds  per  square  foot  of 
area  covered,  and  as  the  area  of  the  roof  is  considerably 
greater  than  this,  owing  to  the  pitch  of  the  roof,  it  is  neces- 


2  Layers  of  felt 


flooring. 


SPAN  OF  GIRDERS  ISO 


— 6-0"Center  to  Center 

FIG.  25. 

sary  to  divide  the  area  covered  by  the  area  of  the  roof  and 
multiply  the  result  by  the  quantities  given  in  the  table. 
For  example,  the  area  of  a  building  covered  by  a  roof  with 
a  span  of  50  feet  is  10,000  square  feet,  and  the  area  of  the 
roof  is  15,000  square  feet;  10, 000 -J- 15,000  =  f,  or  .67,  and, 
since  the  span  of  the  roof  is  50  feet,  according  to  Table  3, 
the  weight  of  the  truss  is  5  pounds  for  each  square  foot  of 
area  covered.  Therefore,  5  X  .  67  =  3. 35  pounds  are  to  be 
added  to  the  weight  of  each  square  foot  of  roof  surface. 


§5  ARCHITECTURAL    ENGINEERING.  31 

EXAMPI.K. — In  Fig.  25,  \\-luit  is  the  total  dead  load  on  the  girder  />' ? 
SOLUTION.— 

Yellow-pine  flooring,  1  inch  thick           =  4     Ib.  per  sq.  ft. 

2  layers  of  felt                                                =  \   Ib.  per  sq.  ft. 

Rough  spruce  flooring,  3  inches  thick  =  (i     Ib.  per  sq.  ft. 

Assume  the  weight  of  the  girder  •         ==  8     Ib.  per  sq.  ft. 

Total  dead  load  of  floor  surface  =  IN.1,  Ib.  per  sq.  ft. 

The  area  of  the  floor  carried  by  the  girder  is  (5  X  IN  =  108  square 
feet.  Then  108  X  W  =  1,998  pounds  is  the  entire  dead  load  upon  the 
girder  />'.  Ans. 

EXAMPL.KS    FOH  PHAC'TIC'K. 

1.  A  2"  X  3"   wrought-iron   bar    is  36.\  inches   long.      What  is  its 
weight?  Ans.  (50. (1(5  11). 

2.  The  outside  diameter  of  a  cast-iron  column  is  10  inches,  and  the 
thickness  of  the  material  composing  the  column  is  J  inch.     What  is  its 
weight  per  foot  of  length  ?  Ans.  (18  Ib. 

'6.  The  wall  of  a  brick  building  was  laid  in  cement  mortar  and  is 
24  inches  thick,  36  feet  high,  and  100  feet  long;  in  it  are  located  20  win- 
dow openings,  2  feet  (5  inches  wide  by  (5  feet  high.  What  is  the  weight 
of  this  wall  ?  Ans.  858,000  11). 

4.  What  is  the  weight  of  a  structural  steel  angle  (5  in.  X  <>  in.  X  '  i". 
X  20  ft.  long  ?  Ans.  390.54  Ib. 

5.  The  roof  of  a  building  is  made  of  No.  20  corrugated  galvanized 
iron,  laid  upon  1-inch  spruce  boarding.     What  is  the  weight  of  the  r<x>f 
covering  per  square  foot  ?  Ans.  4J  Ib. 

6.  What  will  be  the  difference  in  weight  between  a  4-ply  slag  roof, 
laid    upon    3-inch  tongued-and-grooved  yellow-pine    planking,   and  a 
^-inch  slate  roof  laid  upon  2-inch  hemlock  sheathing,  covered  with 
Neponset  roofing  felt,  two  layers  thick  ?  Ans.  3J  Ib. 

7.  The  span  of  a  roof  truss  is  40  feet,  and  its  rise  10  feet.     What 
weight  per  square  foot  of  roof  surface  should  be  assumed  so  as  to  allow 
for  the  weight  of  the  principal  or  roof  truss  ?  Ans.  3.58  Ib. 


LIVE    LOAD. 

42.  Besides  the  dead  load,  which  includes  the  weight  of 
all  the  material  used  in  the  structure  itself,  there  is  a  load 
due  to  the  weight  of  the  people  and  merchandise;  this  load 
is  called  the  live  load.  The  live  load  comprises  people  in 


32  ARCHITECTURAL  ENGINEERING.  §  5 

the  building,  furniture,  movable  stocks  of  goods,  small 
safes,  and  varying  weights  of  any  character.  Large  safes 
and  extremely  heavy  machinery  require  some  special  pro- 
vision, usually  embodied  in  the  construction.  Table  4  gives 
the  live  loads  per  square  foot  recommended  as  good  prac- 
tice in  conservative  building  construction. 


TABLE    4. 


Dwellings 70  Ib. 

Offices   70  Ib. 

Hotels  and  apartment  houses 70  Ib. 

Theaters 120  Ib. 

Churches 120  Ib. 

Ballrooms  and  drill  halls 120  Ib. 

Factories from  150  up. 

Warehouses from  150  to  250  up. 

The  load  of  70  pounds  will  probably  never  be  realized  in 
dwellings ;  but  inasmuch  as  a  city  house  may,  at  some  time, 
be  applied  to  some  purpose  other  than  that  of  a  dwelling,  it 
is  not  generally  advisable  to  use  a  lighter  load.  In  the  case 
of  a  country  house,  a  hotel,  or  a  building  of  light  character, 
where  economy  demands  it,  and  its  actual  use  for  a  long 
time,  for  some  fixed  purpose,  is  almost  certain,  a  live  load 
of  40  pounds  per  square  foot  of  floor  surface  is  ample  for  all 
rooms  not  used  for  public  assembly. 

For  rooms  thus  used,  a  live  load  of  80  pounds  will  be  suf- 
ficient, experience  having  demonstrated  that  a  floor  cannot 
be  crowded  to  more.  If  the  desks  and  chairs  are  fixed,  as  in 
a  schoolroom  or  church,  a  live  load  of  more  than  40  to  50 
pounds  will  never  be  attained.  Retail  stores  should  have 
floors  proportioned  for  a  live  load  of  100  pounds  and 
upwards.  Wholesale  stores,  machine  shops,  etc.  should 
have  the  floors  proportioned  for  a  live  load  of  not  less  than 
150  pounds  per  square  foot. 


§5  ARCHITECTURAL  ENGINEERING.  33 

The  static  load  in  factories  seldom  exceeds  40  to  ~>0  pounds 
per  square  foot  of  floor  surface,  and,  therefore,  in  the 
majority  of  cases,  a  live  load  of  100  pounds,  including  the 
effects  of  vibrations  due  to  moving  machinery,  is  ample. 
The  conservative  rule  is,  in  general,  to  assume  loads  not 
less  than  the  above,  and  to  be  sure  that  the  beams  are  pro- 
portioned so  as  to  avoid  excessive  deflection.  Stiffness  is  a 
factor  as  important  as  strength. 

EXAMPLK. — What  will  be  the  entire  live  load  coming  upon  a  large 
girder  supporting  a  portion  of  a  church  floor,  if  the  floor  area  to  be  sup- 
ported is  600  square  feet  ? 

SOLUTION.— From  the  list  given  in  Table  4,  120  pounds  is  usually 
considered  safe  for  a  live  load  in  a  church.  Therefore,  (iOO  X  120 
=  72,000  pounds  is  the  total  live  load  on  the  girder.  Ans. 


KX  AMI'IYKS   FOK   IMt.VCTICK. 

1.     What    will   be    the    entire    live    load   on    the    floor   of   a   church 
50  ft.  X  120  ft.?  Ans.  720,000  Ib. 

2.  What  live  load  will  a  joist  in  a  city  dwelling  be  required  to  bear, 
the  distance  between  centers  being  14  inches,  and  the  span  of  the  joist, 
20  feet?  Ans.   !,(>:«  Ib. 

3.  A  steel  beam,  supporting  a  portion  of  the  floor  in  an  office  build- 
ing, sustains  an  area  of  80  square  feet.     What   will   be   the   live  load 
coming  upon  the  beam  ?  Ans.  f>,()00  Ib. 


8XOW    AX  I)    AVI  XI)     LOADS. 

43.  In  calculating  the  weight  upon  roofs,  there  are  two 
other  loads  always  to  be  considered  when  obtaining  the 
stresses  on  the  various  members  of  the  truss.  These  are 
snow  and  wind  loads.  Where  the  roof  is  comparatively  flat, 
that  is,  where  the  rise  of  the  roof  is  under  12  inches  per 
foot  of  horizontal  distance,  the  snow  load  is  estimated  at 
12  pounds  per  square  foot;  for  roofs  that  have  a  steep  slope, 
or  a  rise  of  more  than  12  inches  per  foot  of  horizontal  dis- 
tance, it  is  good  practice  to  assume  the  snow  load  to  be  8 
pounds  per  square  foot. 


34 


ARCHITECTURAL  ENGINEERING. 


44.     AVind   pressure   on    roofs   is   always    assumed   as 
acting   normal   (that   is,    perpendicular)   to    the    slope.      In 

Fig.  26,  the  outline  abc 
of  a  roof  is  shown;  the 
force  d  is  normal  to  the 
slope  a  b,  and  represents 
the  assumed  pressure  of 
the  wind  on  the  roof. 
c  The  wind  generally  acts 


in  a  horizontal  direction, 
The  maximum  horizontal 


FIG.  26. 

as  shown  by  the  full  arrow  c. 
pressure  of  the  wind  is  always  considered  to  b'e  40  pounds 
per  square  foot ;  this  pressure  represents  a  wind  velocity  of 
from  80  to  100  miles  per  hour,  which  is  a  violent  hurricane 
in  intensity,  and  as  this  velocity  is  seldom  realized,  and 
never  exceeded  except  in  cyclonic  storms,  the  assumption 
may  be  considered  reasonably 
safe.  The  wind,  blowing  with 
a  horizontal  pressure  of  40 
pounds,  strikes  the  roof  at  an 
angle ;  consequently,  the  pres- 
sure d,  normal  to  the  slope,  is 
considerably  less  than  40 
pounds,  unless  the  slope  of 
the  roof  is  very  steep.  Refer- 
ring to  Figs.  27  and  28,  it  is 
clear  that  the  horizontal  force  c  of  the  wind  on  the  slope  of 
the  roof,  shown  in  Fig.  27,  is  almost  as  intense  as  on  a  vertical 


FIG.  27. 


FIG.  28. 


surface ;  on  the  extremely  flat  roof  in  Fig.  28,  however,  the 
wind  exerts  hardly  any  force  at  all  normal  to  the  slope, 
because  it  strikes  the  slope  at  such  an  acute  angle,  and 


§  5  A RC H I T !•: C T U R A L  E X G I X E E R I X ( i .  :5:> 

therefore  has  a  tendency  to  slide  along  and  otT  it.  The  more 
acute  the  angle  between  the  lines  <•  and  ti,  the  greater  the 
pressure  normal  to  the  slope;  whereas,  the  greater  the 
distance  they  are  apart  or  the  greater  the  angle,  the  less  the 
pressure  normal  to  the  slope,  until  they  form  a  right  angle 
with  each  other,  where  the  pressure  on  the  roof  may  be 
disregarded.  On  the  basis  of  a  horizontal  wind  pressure  of 
40  pounds,  the  pressure  normal  to  the  slope  has  been 
reckoned  by  a  formula  known  as  Uutton's  formula.  This 
formula,  being  trigonometric,  is  not  given  here,  but  results 
derived  from  it  are  given  in  the  following  table: 


TABLE   5. 

•\VIXIJ    PKKSSriJK   NORMAL    TO    TIIK   SI.OT'K   OF   HOOF. 


Rise. 


Pitch, 


Wind 

Anele  of     T-,  J     L1''.         Pressure 
Slope  with    Pr7£rtlon  Normal  to 

Horizontal.  Slope 

to  Span.    jn  pmjmls_ 


4  inches  per  foot 

horizontal.  . 

18°  25'             i 

1C.  8 

6  inches  per  foot 

horizontal.  . 

2G°  33'             i 

2:5.7 

8  inches  per  foot 

horizontal  .  . 

3.V  42' 

a 

2!».  I 

12  inches  per  foot 

horizontal  .  . 

4.")°    ()' 

3(5.1 

16  inches  per  foot 

horizontal.  .      53°    ?' 

A 

38.7 

18  inches  per  foot 

horizontal  .  . 

f>i;°  20' 

1 

31).  3 

24  inches  per  foot 

horizontal  .  . 

03°  27' 

1 

4o.o 

45.  In  order  to  more  fully  explain  Table  5,  refer  to  Fig. 
29.  The  rise  in  the  slope  a  b  is  C  inches  for  every  12  inches 
on  the  horizontal  line  ac\  for  instance,  at  4  feet  from  a  on 
the  horizontal  line  a  c,  the  rise  is  4  times  G  inches,  or  2  feet, 
the  angle  included  between  the  line  of  slope  a  b  and  the 
horizontal  base  line  ac  is  2<>°  33',  and  the  pressure  normal 
to  the  slope,  according  to  Table  5,  is  assumed  at  23.7  pounds 
per  square  foot.  Since  the  rise  at  the  center  is  equal  to 

1-15 


36 


ARCHITECTURAL  ENGINEERING. 


one-half  the  length  of  one-half  the  span,  the  total  rise  is 
one-quarter  of  the  span.  Under  these  conditions,  the  pitch 
of  the  roof,  that  is,  the  ratio  of  the  rise  to  the  span,  is  £, 
and  the  roof  is  said  to  be  \  pitch. 


FIG.  29. 

EXAMPLE. — (a)  What  will  be  the  dead  load  per  square  foot  of  roof 
surface,  on  a  roof  with  a  12-inch  rise,  the  span  of  the  trusses  being  50 
feet,  the  roof  covering  1  inch  white-pine  sheathing,  2  layers  of  Neponset 
roofing  felt,  and  ^-inch  slate  3-inch  lap  ?  (b)  What  will  be  the  wind 
pressure  per  square  foot  normal  to  the  slope  ?  (c)  If  the  roof  trusses 
are  placed  12  feet  apart,  what  will  be  the  entire  dead  load  on  one 
truss  ?  Fig.  30  shows  a  plan  with  elevation  and  detail  section  of  the 
roof. 

SOLUTION. — (a)  By  referring  to  Table  3,  it  is  seen  that  the  approxi- 
mate weight  of  a  roof  truss  with  a  span  of  50  feet  is  5  pounds  for  every 
square  foot  of  area  covered.  It  is  first  necessary  to  obtain  the  length  of 
the  line  of  slope  a  b ;  this  is  done  by  calculating  the  hypotenuse  of  the 
triangle,  or  by  laying  the  figure  out  to  scale  and  measuring.  In  this 
case  it  is  found  that  a  b  measures  about  35  feet  4  inches,  equal  to  35.33 
feet.  The  area  covered  by  the  roof  supported  on  one  truss  is  12  X  50 
=  600  square  feet.  The  area  of  the  roof  supported  by  one  truss  is 
2  X  35.33  X  12  =  847.92  square  feet.  Then,  600  -j-  847.92  =  .70,  which 
means  that  the  weight  of  the  truss  per  square  foot  of  roof  surface  is 
.70  times  5  pounds,  or  5  X  .70  =  3.5  pounds.  The  dead  load  per  square 
foot  of  roof  surface  is,  then,  as  follows: 


ARCHITECTURAL  ENGINEERING. 


Weight  of  supporting  truss ;5..j  Ib.  per  sq.  ft. 

Weight  of  white-pine  sheathing  1  inch  thick  .  '.2.5  Ib.  per  sq.  ft. 
Weight  of  2  layers  of  Xeponset  rooting  paper  .5  Ib.  per  sq.  ft. 
Weight  of  slate  (\  inch  thick) 4.5  Ib.  per  sq.  ft. 


Total 


.  ll.Olb.  per  sq.ft.  Ans. 


The  weight  of  the  purlins  supporting  the  sheathing  has  not  been 
estimated  in  the  above,  it  being  safe  in  this  case  to  assume  that  the 


-  12  Ft. 


50  Ft 
Ele  ra  t  inn  of  Roof 

,"  " 

t  Slate  3  I  tip 

2-layen  of  yeponset-Fc 


I' I  it  H  of  Hoof 


Wood  purlin 


eel  Truns 


Detail  of  Roof  Covering 

Fin.  30. 

weight  used  for  the  principals,  or  trusses,  is  sufficient  to  cover  this  item. 
A  snow  and.  accidental  load  of  12  pounds  per  square  foot  of  roof  surface 
should  also  be  added  to  the  dead  load  to  get  the  entire  vertical  load 
upon  the  roof. 

(6)  The  wind  pressure  normal  to  the  slope  of  this  roof,  according 
to  Table  5,  for  a  one-half  pitch  roof  is  36. 1  pounds,  say  36  pounds  per 
square  foot.  Ans. 


38  ARCHITECTURAL  ENGINEERING.  §5 

(c)  The  area  of  the  roof  supported  by  one  truss  is,  as  previously 
found,  847.92  square  feet,  and  the  dead  load  is  11  pounds  per  square 
foot.  Then,  847.92x11  =  9,327.12  pounds  to  be  supported  by  one 
truss,  not  including  the  snow  load.  Ans. 

46.  Engineering   is  not,   it   must  be  remembered,    an 
exact  science,  the  results  obtained  depending  more  or  less 
upon  the  judgment  and  experience  of  the  designer.     When, 
for  instance,  the  wind  is  blowing  a  hurricane,  snow  never 
lodges  on  a  roof,  the  slates,  shingles,  and  sheathing  being 
themselves,  in  such  a  case,  exposed  to  sudden  removal.    If, 
therefore,  the  full  wind  pressure  be  assumed,  the  snow  load 
may,  in  most  cases,  be  omitted,  especially  if  the  desire  be 
to  build  an  economical  roof.      It  is,  however,  not  well  for 
the  student  to  make  such  assumptions  until  his  experience 
and  judgment  is  sufficiently  developed  to  enable  him  to 
make  true  deductions. 

47.  Careful  designers  sometimes  make  allowance  for 
the  accidental  load  caused  by  a  heavy  body  falling  upon 
the  floor,  or  by  a  mass  of  snow  dropping  from  one  roof  to 
another.      But  this  load  may  usually  be  ignored,  because  it 
is  taken  care  of  in  the  factor  of  safety,  within  the  limit  of 
which  every  member  in  a  structure  is  designed. 


EXAMPLES  FOR  PRACTICE. 

1.  The  area  of  one  slope  of  a  one-half  pitch  roof  is  800  square  feet. 
What  is  the  entire  pressure  upon  the  slope  of  the  roof,  provided  the 
maximum  horizontal  wind  pressure  is  taken  at  40  pounds  per  square 
foot  ?  Ans.  28,800  Ib. 

2.  In  a  quarter-pitch  roof  the  trusses  are  20  feet  apart,  and  the  length 
of  the  roof  slope  is  40  feet.     What  wind  load  is  there  upon  each  roof 
truss,  if  the  horizontal  pressure  is  40  pounds  per  square  foot  ? 

Ans.  18,960  Ib. 

3.  The  purlins  supporting  a  f-pitch  roof  are  placed  6  feet  apart,  and 
the  trusses  are  12  feet  from  center  to  center.     What  is  the  maximum 
load  due  to  the  wind  upon  each  purlin,  providing  the  greatest  hori- 
zontal pressure  is  40  .pounds  per  square  foot  ?  Ans.  2,830  Ib. 


ARCHITECTURAL  ENGINEERING.  3!) 


STRESSES  AXD   STRAIXS. 

DEFINITIONS. 

48.  It  has  been  shown  that  the  weight  of  the  materials 
composing  a  building  and  its  contents  produces  forces  that 
must  be  resisted  by  the  different  members  of  the  structure; 
the  action  of  these  forces  has  a  tendency  to  change  the 
relative  position  of  the  particles  composing  the  members, 
and  this  tendency  is,  in  turn,  resisted  by  the  cohesive  force 
.  in  the  materials,  which  acts  to  hold  the  particles  together. 

The  internal  resistance  with  which  the  force  of  cohesion 
opposes  the  tendency  of  an  external  force  to  change  the 
relative  position  of  the  particles  of  any  body  subjected  to  a 
load  is  called  a  stress.  Or,  stress  may  be  defined  as  the 
load  per  square  inch  which  produces  a  fractional  alteration 
in  the  form  of  a  body,  and  this  fractional  alteration  of  form 
is  called  the  strain. 

4J).  In  accordance  with  the  direction  in  which  the  forces 
act  with  reference  to  a  body,  the  stress  produced  may  be 
either  tensile,  compresslve,  or  shearing. 

50.  Tensile   stress   is    the    effect    produced  when   the 
external  forces  act  in  such   a  direction   that  they  tend  to 
stretch  a  body;  that  is,  to  pull  the  particles  away  from  each 
other.    A  rope  by  which  a  weight  is  suspended  is  an  example 
of  a  body  subjected  to  a  tensile  stress. 

51.  Compressive  stress  is   the   effect  produced  when 
the  tendency  of  the  forces  is  to  compress  the  body  or  to 
push  the  particles  closer  together.     A  post  or  the  column  of 
a  building  is  an  example  of  a  body  subjected  to  a  compress- 
ive  stress. 

52.  fciheariiitf  stress  is  the  effect  produced  when  the 
forces  act  as  in  a  shear,  so  as  to  produce  a  tendency  for  the 
particles  in  one  section  of  a  body  to  slide  over  the  particles 
of  the  adjacent  section.     When  a  steel  plate  is  acted  on  by 
a  punch  or  the  knives  of  a  shear,  or  where  a  load  acts  on  a 


40 


ARCHITECTURAL  ENGINEERING. 


beam,  as  shown  in  Fig.  31,  the  plate  or  beam  is  subjected 
to  a  shearing  stress. 


Support 


Weight 


Support. 


FIG.  31. 


53.  When  a  beam  is  loaded  in  such  a  manner  that  there 
is  in  it  a  tendency  to  bend,  as  shown  in  Fig.  32,  it  is  sub- 
jected to  a  transverse,  or  bending,  stress.  There  is,  in 
this  case,  a  combination  of  the  three  above  mentioned 


Support 


FIG.  32. 


stresses    (tension,    compression,    and    shear)    in    different 
parts  of  the  beam. 

54.  There  is  still  another  type  of  stress  called  torsion, 
which,  however,  is  comparatively  seldom  met  with  in  build- 
ing construction.  An  example  of  a  body  subjected  to  a 
torsional,  or  twisting,  stress  is  a  shaft  which  carries  two 
pulleys,  one  of  which  "is  acted  on  by  the  driving  force  of  a 
belt.  The  force  transmitted  from  the  driven  pulley  through 
the  shaft  produces  a  tendency  to  twist  the  shaft.  The  effect 
of  this  twisting  action  is  a  tendency  to  slide  the  particles  in 


§  5  ARCHITECTURAL  ENGINEERING.  41 

any  two  adjacent  sections  of  the  shaft  over  each  other. 
Torsion  may,  consequently,  be  included  under  the  head  of 
shearing  stress. 

55.  The    unit    stress   (called,    also,    the    intensity    of 

stress)  is  the  name  given  to  the  stress  per  unit  of  area; 
or,  it  is  the  total  stress  divided  by  the  area  of  the  cross- 
section.  Thus,  if  a  weight  of  1,000  pounds  is  supported  by 
an  iron  rod  whose  area  is  4  square  inches,  the  unit  stress  is 
!_o^.o.  _  250  pounds  per  square  inch.  If,  with  the  same 
load,  the  area  is  .V  square  inch,  the  unit  stress  is  —  ?-°-  =  2,000 
pounds  per  square  inch. 

Let      P  —  total  stress  in  pounds; 

A   =  area  of  cross-section  in  square  inches ; 
J>    =  unit  stress  in  pounds  per  square  inch. 

Then,  S  =    ^,  or  P  =  A  S.  (1.) 

That  is,  the  total  stress  is  equal  to  the  area  of  the  seetion 
multiplied  by  the  unit  stress. 

EXAMPLE. — An  iron  rod,  2  inches  in  diameter,  sustains   a  load  of 
90,000  pounds ;  what  is  the  unit  stress? 
SOLUTION. — Using  formula  1, 

P  90,000 

o  =  —j  =     ,2  ---nuKA.  =  28,047.8  Ib.  per  sq.  in.     Ans. 

56.  When  a  body  is  stretched,  shortened,  or  in  any  way 
deformed  through  the  action  of  a  force,  the  deformation  is 
called  a  strain.     Thus,  if  the  rod  before  mentioned  had 
been  elongated  y1^  inch  by  the  load   of   1,000  pounds,  the 
strain  would  have  been  y1^  inch.     Within  certain  limits,  to 
be  given  hereafter,  strains  are   proportional  to  the  stresses 
producing  them. 

57.  The  unit  strain  is  the  strain  per  unit  of  length  or 
of  area,  but  is  usually  taken  per  unit  of  length  and  called 
the  elongation  per  unit  of  length.      If  we  consider  the  unit 
of  length  as  1    inch,   the   unit    strain   is   equal  to  the  total 
strain  divided  by  the  length  of  the  body  in  inches. 


ARCHITECTURAL  ENGINEERING.  §  5 

Let  /  =  length  of  body  in  inches; 

e  =  elongation  in  inches; 
s  =  unit  strain. 

Then,  s  =    ,  or  c  =  Is.  (2.) 


EXAMPLES  FOR    PRACTICE. 

1.  A  wrought-iron  tension  member  in  a  roof  truss  has  a  load  upon 
it  of  27,000  pounds.     If  it  has  been  figured  to  sustain  a  working  stress 
of  6,000  pounds,  what  will  be  its  diameter  ?  Ans.  2|  in.,  nearly. 

2.  The  sectional  dimensions  of  a  wooden  compression  member  are 
8  in.  X  10  in. ;  if  the  load  upon  it  is  64,000  pounds,  what  is  the  unit 
stress  upon  the  material  ?  Ans.  800  Ib. 

3.  What  is  the  total  load  upon  a  hollow  cast-iron  column  10  inches 
outside  diameter;   the  thickness  of  the  metal  is  1  inch,  and  the  unit 
stress  upon  the  column  is  20,000  pounds  ?  Ans.   565,600  Ib. 

4.  Two  steel  angles  form  the  tension  member  in  a  roof  truss  and 
have  a  combined  sectional  area  of  3}  square  inches;  they  are  subjected 
to  excessive  stress,  and  are  stretched  \  inch.     What  is  the  unit  strain 
in  them  if  they  are  10  feet  long  ?  Ans.  .00208  in. 


STREXGTH  -OF    BUILDIXG    MATERIALS. 

58.  The   ultimate    strength  of  any   material  is  that 
unit  stress  which  is  just  sufficient  to  break  it. 

59.  The  ultimate  elongation  is  the  total  elongation  pro- 
duced in  a  unit  of  length  of  the  material  having  a  unit  of  area, 
by  a  stress  equal  to  the  ultimate  strength  of  the  material. 

GO.  Modulus  of  Rupture. — The  fibers  in  a  beam  sub- 
jected to  transverse  stresses  are  either  in  compression  or 
tension,  depending  whether  they  are  above  or  below  the 
neutral  axis.  It  has,  however,  been  determined  that  the 
strength  of  the  extreme  fibers  in  a  beam  neither  agree  with 
their  compressive  or  tensile  strength.  Hence,  in  beams  of 
uniform  cross-section  above  and  below  the  neutral  axis  it  is 
usual  to  use  a  constant,  which  has  been  determined  by 
actual  tests.  This  constant  is  called  the  modulus  of  rupture, 
and  is  generally  expressed  in  pounds  per  square  inch. 


§  5 


ARCHITECTURAL  ENGINEERING. 


61.  Table  G  gives  values  of  the  strength  of  building 
materials  commonly  used,  when  subjected  to  different 
stresses. 


TABIVE    G. 
STRENGTH   OF    MATERIALS,    IN    POUNDS    PER   SQUARE    INCH. 


Material. 

mate  Tensile. 

mate  Compres- 
Parallel  to  the 
Grain. 

rable  Compres- 
Perpendicular 
j  the  Grain. 

Ulti 
Shea 

0       JH 

2  2 

Tiate 
ring. 

oji 

o 

K-i 

0 

§ 

II 

£  o 

II 

£3 

0 

White  pine  .  . 

6,000 
4,000 
6,000 
8,000 
10,000 
50,000 
48,000 
60,000 

3,000 
2,000 
3,000 
4,400 
3,  GOO 
44,000 

250 
250 
300 
600 
700 

300 
250 
300 
400 
GOO 
44,000 

2,  500 
2,500 
3,000 
4,500 
5,000 

4,  800 
3,600 
4,800 
7,300 
6,000 
48,000 

Hemlock  
Spruce  

Yellow  pine  .... 
Oak  

Wrought  iron  .  . 
Shape  iron  . 

Structural  steel  . 

to 

65,000 

52,000 

52,000 

60,000 

Allowable, 

5,000 

Cast  iron  
Granite  

18,000 

81,000 
15,000 
7,000 
5,000 

- 

25,000 

45,000 
1,800 
1,500 
1  200 
to  700 

Limestone.  .  . 

Sandstone    

Good  sandstone. 

10,000 

1  700 

NOTE. — The  terms  "parallel  to  the  grain"  and  "  perpendicular  to 
the  grain"  apply  to  wood  only. 

62.  The  values  given  above,  under  their  several  head- 
ings, being  conservative,  are  on  the  safe  side.  The  column 
in  the  table  headed  "Ultimate  Compression  Parallel  to  the 


ARCHITECTURAL  ENGINEERING. 


Grain  "  will  be  found  useful  in  computing-  the  strength  of 
columns.  The  values  in  the  column  headed  "Allowable 
Compression  Perpendicular  to  the  Grain  "  are  used  in  cases 


FIG.  33. 

similar  to  Fig.  33,  and  are  such  as  will  not  produce  an 
indenture  of  more  than  y^  of  an  inch  in  the  surface  of  the 
timber,  a  value  well  within  the  safe  limit.  The  values 
under  the  heading  of  "Ultimate  Shearing  Parallel  to  the 
Grain  "  are  used  in  computing  the  strength  of  the  end  of 


FIG.  34. 

the  tie-beam  at  the  heel  of  the  main  rafter  in  a  roof  truss, 
as  shown  in  Fig.  34.  The  tendency  is  to  shear  off  the  piece 
/r  parallel  to  the  grain  along  the  line  a  b.  The  figures  in 
the  column  headed  "Modulus  of  Rupture"  are  the  con- 
stants, or  values,  used  when  computing  the  strength  of 
beams.  When  a  simple  beam  breaks,  the  fibers  at  the  top 


ARCHITECTURAL  ENGINEERING. 


side  are  in  compression,    and  those  at  the  bottom   side  in 

tension,  as  shown  in  Fig.   35.      By  actual  tests,  it  has  been 

found  that  though  some  of  the  different  fibers  of  materials 

under  transverse  stresses  are  in  compression  and  some  in 

tension,   the   ultimate  resistance  of  the   material  does  not 

agree  with  the  ultimate  resistances  of  the  fibers  to  either 

tension  or  compression.      Though  many  attempts  have  been 

made  to  account  for  it,  this  fact  remains ;  hence,  it  becomes 

necessary  to  obtain  some  constant,  or  value,   more  closely 

agreeing     with    the 

strength  of   materials 

under     transverse 

stresses.     It  is  usual, 

therefore,    where    the 

cross-section     of     the 

beam   is   uniform,    to  FIG.  x>. 

obtain,  by  actual   tests,    the  constants,   or  values,  for  each 

material,  and  these  values  are  called  the  modulus  of  rupture 

and  are  generally  expressed  in  pounds  per  square  inch. 

EXAMPLE  1. — What  pull  will  be  required  to  break  a  2-inch  diameter 
rod  of  wrought  iron  ? 

SOLUTION. — The  area  of  the  rod  is  equal  to  the  area  of  a  2-inch 
diameter  circle,  which  is  22  X  .7854  =  3.14  square  inches;  the  ultimate 
tensile,  or  breaking,  strength  of  wrought  iron,  according  to  Table  6,  is 
about  50,000  pounds  per  square  inch.  Therefore,  the  ultimate  strength 
of  the  rod  in  question  is  about  3.14  X  50,000  =  157,000  pounds.  Ans. 

EXAMPLE  2. — What  length  of  wrought-iron  bar,  if  hung  by  one  end, 
will  break  of  its  own  weight  ? 

SOLUTION. — Assume  any  size  of  bar ;  say  1^  inches  in  diameter.  The 
area  of  this  bar  is  .99  square  inch,  which  may,  for  convenience,  be 
called  1  square  inch.  Wrought  iron,  according  to  Table  1,  weighs  .277 
pound  per  cubic  inch.  Now,  as  there  is  just  1  cubic  inch  in  each 
lineal  inch  in  the  rod,  a  length  of  1  foot  weighs  .277X12  =  3.32 
pounds.  The  tensile  strength  of  wrought  iron  being  50,000  pounds 
per  square  inch,  and  1  foot  of  its  length  weighing  3.32  pounds,  the 

length  of  rod  required  is  —      r  =  15,060  feet.     Ans. 

o.  0/3 

EXAMPLE  3. — In  Fig.  36  is  shown  the  splice  of  a  tie-beam  in  a  wood 
roof  truss  composed  of  yellow  pine.  What  is  the  strength  of  the 
splice,  disregarding  the  bolts  a,  a  entirely  ? 


40 


ARCHITECTURAL  ENGINEERING. 


SOLUTION. — The  strength  of  the  splice  depends  on  the  tensile  strength 
of  the  wood  at  the  net  section  ef  and  upon  the  tensile  strength  of  the 
net  section  of  the  two  splice  plates.  It  also  depends  upon  the  tendency 
of  the  splice  plates  to  srfear  along  the  lines  s  t  and  s'  /',  and  upon  the 
tendency  of  the  tie  to  shear  along  the  line  m  n  and  in'  n' .  Assume  the 
areas  of  the  net  sections  to  be  sufficient  to  make  their  strength  greater 
than  that  of  the  sections  which  will  fail  by  shearing ;  then  referring  to 
Fig.  36,  it  will  be  seen  that  the  line  of  shear  on  the  splice  plate  at  s  t 


a  a, 


,. 

ff 

*' 

:                    |w,'     !  <^ 

,    - 

"°--          -r=H 

-~  ~~ 

_  __. 

—  l»f~-  V  

" 

^ 

i'~"  j 

::      /                      -  ^-~ 

— 

8 

FIG.  30. 

and  s'  /'  is  longer  than  that  of  the  tie  member  at  m  n  and  m'  n' ;  there- 
fore, the  strength  of  the  tie  along  the  line  ;//  n  and  tn'  n'  only  need  be 
considered  in  computing  the  strength  of  the  splice.  The  pieces  o  and  o' 
tend  to  slide  or  shear  off  of  the  main  tie  along  the  lines  m  n  and  m'  n'. 
The  area  along  these  lines  is  12  X  10  X  2  =  240  square  inches.  Refer- 
ring to  Table  6  under  the  column  headed  "  Ultimate  Shear  Parallel  to 
the  Grain,"  the  value  for  yellow  pine  is  found  to  be  400  pounds  per 
square  inch.  Hence,  240  X  400  =  96,000  pounds,  the  ultimate  strength 
of  the  splice,  disregarding  the  bolts  a,  a. 

63.     The  factoi-  of  safety,  or,  as  some  call  it,  the  safety 

factor,  is  the  ratio  of  the  breaking  strength  of  the  structure 
to  the  load  which,  under  usual  conditions,  it  is  called  upon 
to  earn-.  Suppose  the  load  required  to  break,  dismember, 
or  crush  a  structure  is  5,000  pounds,  and  the  load  it  is  called 
upon  to  carry  is  1,000  pounds,  then  the  factor  of  safety  may 
be  obtained  by  dividing  the  5,000  pounds  by  the  1,000 
pounds,  or  %%%%  =  5,  the  factor  of  safety  in  this  structure. 

The  safety  factor  depends  upon  the  conditions,  circum- 
stances, or  materials  used ;  in  other  words,  it  is  the  factor 
of  ignorance.  When  a  piece  of  steel,  wood,  or  cast  iron  is 
used  in  a  building,  the  engineer  does  not  know  the  exact 
strength  of  that  particular  piece  of  steel,  wood,  or  cast  iron. 
From  his  own  experience,  and  that  of  others,  he  knows  the 
approximate  tensile  strength  of  structural  steel  to  be  60,000 


§  5  ARCHITECTURAL  ENGINEERING.  47 

pounds  per  square  inch,  and  that  it  varies  more  or  less  from 
this  value.  In  regard  to  timber,  the  uncertainty  is  much 
greater,  because  of  knots,  shakes,  and  interior  rot,  not 
always  evident  on  the  surface.  Cast  iron  is  even  more 
unreliable,  on  account  of  almost  indeterminable  blowholes, 
flaws,  and  imperfections  in  the  castings. 

64.  Deterioration. — Another  factor  to  be  considered  is 
deterioration  in  the  material,  due  to  various  causes.  In 
metals  there  is  corrosion  on  account  of  moisture  and  gases 
in  the  atmosphere,  especially  noticeable  in  the  steel  trusses 
over  railroad  sheds,  where  the  sulphur  fumes  from  the  stacks 
of  the  locomotives  unite  with  the  moisture  in  the  air,  form- 
ing free  sulphuric  acid,  which  attacks  the  steel  vigorously, 
and  demands  constant  painting,  to  prevent  its  entire  destruc- 
tion. Wood  is  subject  to  decay  from  either  dry  or  wet  rot, 
caused  by  local  conditions;  it  may,  like  iron  and  steel,  be 
subjected  to  fatigue,  produced  by  constant  stress  due  to  the 
load  it  may  have  to  sustain.  Cast  iron  does  not  deteriorate 
to  any  great  extent,  its  corrosion  not  being  as  rapid,  possibly, 
as  that  of  steel  or  wrought  iron.  There  are,  however, 
internal  strains  produced  in  cast  iron  by  the  irregular  cool- 
ing of  the  metal  in  the  mold.  Castings  under  the  slightest 
blow  will  sometimes,  owing  to  these  internal  stresses,  snap 
and  break  in  a  number  of  places. 

These  reasons  are,  in  truth,  sufficiently  cogent  to  require 
the  factor  of  safety  now  adopted  in  all  engineering  work. 
Table  7  gives  the  factor  of  safety  recognized  by  conservative 
and  judicious  constructors,  for  various  materials. 


TABI/E    7. 

SAFETY  FACTORS  FOR  DIFFERENT  MATERIALS  ITSED 
IX  CONSTRUCTION. 


Structural  steel  and  wrought  iron '}  to  4. 

Wood   4  to  5. 

Cast  iron G  to  10. 

Stone    .  .  10  at  least. 


48  ARCHITECTURAL  ENGINEERING.  §5 

In  the  above  table  the  factor  of  safety  generally  used  for 
structural  steel  is  3  to  4,  which  simply  means  that  the  steel 
structure  should  not  break  until  it  bears  a  load  3  or  4  times 
greater  than  it  is  expected  to  carry. 

EXAMPLE. — If  the  breaking  strength  of  a  cast-iron  column  is  200,000 
pounds,  what  safe  load  will  the  column  sustain  if  a  factor  of  safety  of  6 
is  used  ? 

SOLUTION.—    200,000 -f- 6  =  33,333  pounds.     Ans. 

EXAMPLES  FOR  PRACTICE. 

1.  Providing  a  factor  of  3  is  adopted,  what  will  be  the  safe  working 
stress  on  a  2-inch  diameter  tension  rod  of  structural  steel  ? 

Ans.  62,800  Ib. 

2.  The  pull  upon  a  2-inch  eyebolt,  passing  through  a  piece  of  yellow 
pine,  is  40,000  pounds.     What  should  be  the  diameter  of  the  washer,  if 
the  bolt  hole  through  it  is  2J-  inches  in  diameter  ?  Ans.  9|  in. 

3.  What  will  be  the  crushing  strength  of  a  granite  capstone  24 
inches  square  ?  Ans.  8,640,000  Ib. 

4.  The  bottom  of  the  notch  in  a  yellow-pine  timber  10  inches  wide 
and  12  inches  deep,  forming  the  tie  member  in  a  roof  truss,  is  18  inches 
from  the  end.    What  resistance  will  the  end  of  the  tie  offer  to  the  thrust 
of  the  rafter  ?  Ans.  72,000  Ib. 

5.  A  short  block  of  yellow  pine,  10  in.  X  10  in.  in  section,  standing 
on  end,  supports  50,000  pounds.  What  is  its  factor  of  safety?    Ans.  8.8. 


FOUNDATIONS. 


STRENGTH    OF    FOUNDATION    MATERIALS. 

65.  It  is  useful,  before  considering  the  strength  of 
columns  or  compression  members  in  buildings,  to  take  up 
the  compressive  strength  of  brickwork  and  masonry,  these 
being  the  mediums  by  which  the  columns  or  posts  in  a  build- 
ing transmit  their  compressive  resistance  to  the  ground. 
As  the  bearing  strength  of  the  soil,  that  is,  its  capacity  to 
support  a  greater  or  less  load,  determines  the  spread  of  the 
foundation,  we  here  submit  bearing  values  for  brickwork 
and  stonework  and  the  several  kinds  of  soils  likely  to  be 
encountered  in  building  operations. 

Table  8  gives  the  conservative  bearing  values  of  brick- 
work, masonry,  and  soils. 


ARCHITECTURAL  ENGINEERING.  -4!) 


TABKE    S. 

THE   SAFE   BEARING   VALVES   OF  BRICKWORK,  MASONRY, 
ANL>   SOILS. 


BRICKWORK. 

Brickwork,  hard  bricks,  dried  lime 

mortar ,  .  100  Ib.  per  sq.  in. 

Brickwork,  hard  bricks,  dried  Port- 
land cement  mortar 200  Ib.  per  sq.  in. 

Brickwork,  hard  bricks,  dried  Rosen- 
dale  cement  mortar 150  Ib.  per  sq.  in. 

MASONRY. 

Granite,  capstone,  in  lime  mortar . .    TOO  Ib.  per  sq.  in. 
Sandstone,  capstone,  in  lime  mortar  350  Ib.  per  sq.  in. 

Bluestone  (a  sandstone) 500  to  TOO  Ib.  per  sq.  in. 

Limestone,  capstone,  in  lime  mortar  500  Ib.  per  sq.  in. 
Granite,    square   stone  masonry,  in 

lime  mortar 350  Ib.  per  sq.  in. 

Sandstone,   square    stone    masonry, 

in  lime  mortar 1T5  Ib.  per  sq.  in. 

Limestone,  square  stone  masonry,  in 

lime   mortar 250  Ib.  per  sq.  in. 

Rubble  masonry,  in  lime  mortar.  . .      80  Ib.  per  sq.  in. 
Rubble      masonry,       in      Portland 

cement   mortar 150  Ib.  per  sq.  in. 

Concrete,    Portland     cement    (1    of 

cement,  2  of   sand,   5    of    broken 

stone)   150  Ib.  per  sq.  in. 

SOIL. 

Rock  foundation 20  tons  per  sq.  ft. 

Gravel  and  sand  (compact) 0  to    10  tons  per  sq.  ft. 

Gravel  and  sand  (mixed    with    dry 

clay) 4  to      0  tons  per  sq.  ft. 

Stiff  clay,  blue  clay 2.5  tons  per  sq.  ft. 

Chicago   clay 1  to  1.5  tons  per  sq.  ft. 

In  the  values  given  for  masonry,  the  height  of  the  wall 
should  not  be  over  16  times  the  thickness. 


50 


ARCHITECTURAL  ENGINEERING. 


DESIGN    OF    FOUNDATIONS. 

66.  Proper  designs  for  foundations  are  of  the  utmost 
importance.  The  maximum  load  carried  by  the  foundation 
must  first  be  obtained.  The  loads  to  be  considered  in 
buildings  are  of  two  kinds,  the  dead  and  live  loads,  previ- 
ously explained.  The  live  load  is  variable.  In  office  biiild- 
ings,  parts  of  the  floor  may  be  loaded  to  their  full  capacity, 
but  the  probability  of  the  entire  structure  being  so  loaded 
is  remote.  In  breweries,  storage  warehouses,  factories,  and 
buildings  for  similar  purposes,  all  the  floors  may  be,  how- 
ever, fully  loaded.  The  maximum  of  both  dead  and  live 
loads  must  be  considered  and  the  area  of  the  footings  for 
the  foundations  such  that  the  greatest  pressure  on  different 
soils  does  not  exceed  that  given  in  Table  8. 

In  designing  foundation  footings  or  piers  for  the  sup- 
port of  columns,  certain  proportions  are  considered  best, 
and  are  therefore  generally  adopted.  Fig.  37  presents  a 


7 

\     I 

cit. 

\" 

tt  Iron  Column. 

-22— 

\    II 

Cap  Stdn,e& 

T 

i      i 

Never 
•more  than. 

1             1 

1 

/"                                                                                // 

fJ  —  LBr  *  ckwork^_T_ 

x   JC''.1      '       '                     ' 

W   *   ^    1 

~nl  <m  ^/*e  horizontal. 

6     '     '    i      '    . 

-±    ft 

-.  ••:  ^S^^^^M^^s^ms^^^^^^^mm 

FIG.  37. 

diagram  of  a  properly  proportioned  foundation  pier  for  a 
column.  The  thickness  of  the  capstone  should  be  about 
one-half  the  width  of  the  side.  The  concrete  offset,  as 
shown  at  b,  should  never  be  more  than  8  inches.  The  layer 
of  concrete  may  be  from  12  to  18  inches  thick,  and  the 
batter  of  the  main  body  of  the  foundation  1  to  2 ;  that  is, 
for  every  foot  of  rise  it  should  incline  6  inches  horizontally. 


§  5  ARCHITECTURAL  ENGINEERING.  51 

67.  As  an  example  of  the  method  of  designing'  a  foun- 
dation pier  for  a  column,  let  us  assume  that  the  load  upon 
a  cast-iron  base  supporting  a  wood  column  is  200,000  pounds, 
and  that  we  are  required  to  design  a  foundation  pier  for  this 
column,  the  pier  being  made  of  brickwork  in  cement  mortar 
with  concrete  base  and  granite  cap. 

The  soil  under  the  pier,  being  compact  gravel  and  sand,  can 
safely  sustain  G  tons  per  square  foot.  The  load  upon  the  soil, 
transmitted  through  the  pier,  is  200,000  pounds,  or  100  tons. 


'I  ., 

II 

Wood 

J   ^ 

Column. 

-  C«s^  J»Y»«  -B««e. 

T 

N 

'-"'    *>  '  e  " 
&     o 

Gh-anite 

Slope  2tOl 


FIG.  38. 

Assuming  a  maximum  load  on  the  soil  of  only  one-half  of 
its  safe  bearing  value,  we  have  100-4-3  =  33£  square  feet 
that  the  base  of  the  concrete  must  cover.  Therefore,  the 
concrete  base  should  be  about  5  feet  0  inches  square.  Next, 
it  is  required  to  find  how  large  the  brick  pier  should  be  at 
the  top,  or  at  the  surface  marked  a,  in  Fig.  38.  The  bearing 
value  of  brickwork  in  Portland  cement  mortar,  according  to 
Table  8,  is  200  pounds  per  square  inch.  The  load  coming 
upon  the  pier  is  200,000  pounds.  Then,  200, 000 -=-200 
=  1,000  square  inches,  the  required  area  at  the  surface  a; 
l,000-f-144  =  6.93  square  feet.  The  upper  surface  a  of  the 
brickwork  being  a  square,  the  length  of  one  side  should  be 
4/6.93,  or  about  2  feet  8  inches.  It  is  now  required  to 

1-16 


52  ARCHITECTURAL  ENGINEERING.  §  5 

determine  the  area  of  the  brickwork  at  the  bottom,  or  where 
it  rests  upon  the  concrete  base.  The  concrete  base,  according 
to  Table  8,  will  with  safety  support  150  pounds  per  square 
inch.  As  the  pressure  upon  it  is  200,000  pounds,  the  area 
in  square  inches  at  this  point  must  be  200,000-1-150  =  1,333 
square  inches,  or  9.25  square  feet.  A  square  whose  sides 
are  3  feet  1  inch  has  an  area  of  9.50  square  feet,  which 
would,  in  theory,  be  about  the  area  of  the  brick  base  next 
to  the  concrete.  In  the  case  before  vis,  represented  by  Fig. 
38,  it  happens,  however,  that  the  area  of  the  concrete  base 
required  is  5  feet  9  inches  on  a  side,  while  the  greatest  limit 
to  which  it  may  extend  beyond  the  brick  pier  is,  according 
to  good  practice,  abovit  8  inches,  due  to  its  liability  to  break 
off  at  the  line  dc;  adoption  of  the  theoretical  area  of  the 
pier  at  this  point  is,  therefore,  inconsistent  with  good  prac- 
tice, the  edge  of  the  brick  pier  being  carried  out  to  within 
8  inches  of  the  edge  of  the  concrete,  regardless  of  dimen- 
sions obtained  in  calculating  the  required  area  for  brick 
piers  bearing  upon  the  concrete  bases.  As  the  granite  cap 
bears  upon  the  brickwork  at  the  surface  a,  its  area  is  gov- 
erned by  the  bearing  strength  of  the  brickwork,  and  is 
required  to  be,  as  previously  found,  2  feet  8  inches  square. 
The  area  of  the  cast-iron  base  is  governed  by  the  permissible 
unit  pressure  on  the  granite  capstone,  which  is  700  pounds 
per  square  inch.  Therefore,  200,000-^-700  =  285  square 
inches  required  to  be  covered,  which  means  a  cast-iron  base 
about  17  inches  square.  The  distance  that  the  capstone 
extends  beyond  the  base  should  not  be  over  one-half  of  the 
thickness  of  the  capstone.  The  capstone  being  in  thickness 
one-half  the  width  of  the  side,  its  thickness  in  this  is  £  of 
2  feet  8  inches  =  1  foot  4  inches,  or  1C  inches.  The  dis- 
tance c  in  this  cap  is,  then,  with  safety  placed  at  £  of  16,  or 
8  inches.  The  cast-iron  base  being  17  inches  square,  and 
the  cap  2  feet  8  inches,  or  32  inches  square,  the  distance  c, 
in  this  case,  would  be  32-17  =  15,  which  divided  by  2 
=  7£  inches,  a  figure  well  within  the  limit. 

Fig.  38  shows  this  pier  foundation  drawn  to  scale,  accord- 
ing to  the  figures  reached  by  the  above  calculation,  in  which 


ARCHITECTURAL  ENGINEERING. 


the  weight  of  the  pier  was  not  considered,  such  exactitude 
not  being  generally  looked  for  in  ordinary  building  con- 
struction. 

EXAMPLES   FOR  PRACTICE. 

1.  "What  safe  load,  uniformly  applied,  will  a  3'  X  3'  brick  pier  laid 
in  Rosendale  cement  mortar  sustain,  the  height  of  the  pier  being  10 
feet?  Ans.   259,200  Ib. 

2.  The  load  transmitted  by  a  cast-iron  column  to  a  foundation  pier 
is  200,000  pounds.     What  should  be  the  size  of  the  base  of  the  pier, 
providing  the  soil  is  compact  gravel  and  sand  ?    Ans.  4  ft.  I  in.  square. 

3.  "What  size  of  granite  capstone,  resting  upon  a  brick  pier  laid  in 
Portland  cement  mortar,  will  be  required   under  a  cast-iron  column 
transmitting  a  load  of  22,000  pounds  ?  Ans.  10^  in.  square. 

4.  A  brick  pier  rests  upon  stiff  clay.     If  the  footing  is  5  feet  square, 
what  load  will  it  safelv  sustain  ?  Ans.  120,000  Ib. 


COLUMXS. 

SHORT    COT.UMXS. 

68.     The  column  may,  in  its  first  stage  of  development, 

be  considered  a  cubical  or 
rectangular  block,  shown 
in  Fig.  39.  As  long  as  the 
column  does  not  exceed 
from  G  to  about  10  times 
the  width  of  the  least  side, 
the  load  it  can  safely 
carry  may  be  estimated 
by  multiplying  its  sec- 
tional area  in  square 
inches  by  the  safe  resist- 
ance to  compression  of 
the  material  parallel  to 
the  grain.  To  get  the 
safe  resistance  of  a  short 
column  or  block,  divide 

FlG.  39.  the  ultimate  resistance  to 

compression  of    the  material   parallel  to  the  grain  by  the 


I 

L 


54 


ARCHITECTURAL  ENGINEERING. 


factor  of  safety,  which,  for  wood  columns,  according  to 
Table  7,  is  from  4  to  5,  though  it  may  be,  in  some  instances, 
good  practice  to  use  G.  It  is  all  a  matter  of  judgment,  gov- 
erned by  the  conditions  to  be  met  with.  After  having 
obtained  the  safe  resistance  of  the  material  to  compression, 
multiply  by  the  sectional  area  of  the  column  in  square 
inches,  and  the  result  will  be  the  safe  resistance  of  the  short 
column  to  compression. 

EXAMPLK. — What  safe  load  will  a  short  yellow-pine  block,  12  inches 
square  and  6  feet  long,  standing  on  end  support,  using  a  safety  factor  of  5  ? 

SOLUTION. — According  to  Table  6,  the  ultimate  compression  of  yellow 
pine  parallel  to  the  grain,  per  square  inch,  is  4,400  pounds;  using  a 
factor  of  safety  of  5,  the  safe  load  per  square  inch  on  this  short  column 
would  be  4, 400 -f- 5  =  880  pounds  per  square  inch.  The  area  of  the 
column  is  12  in.  X  12  in.  =  144  square  inches,  and,  therefore,  144  X  880 
=  126,720  pounds,  the  safe  resistance  to  compression  of  the  short 
column.  Ans. 


T.OXG    COL.UMXS. 

69.  The  statements  of  the  preceding  paragraphs  do  not 
j  apply  to  columns  of  over  10  times  the 
|  width  of  the  least  side.  When  long  col- 
umns are  under  compression,  and  not 
secured  against  yielding  sideways,  it  is 
evident  they  are  liable  to  bend  before 
breaking.  To  ascertain  the  exact  stress  in 
such  pieces  is,  sometimes,  quite  difficult. 
Hence,  we  must  have  a  formula  making 
due  allowance  for  this  tendency  in  the 
column  to  bend,  or  to  split  and  spread 
from  the  center,  as  sho\vn  in  Fig.  40. 

7O.     Formula  for  Wood  Columns. — 

The  formula  mostly  used  for  long,  square, 
or  rectangular  wood  columns,  with  square 
ends,  is  the  following,  deduced  from 
elaborate  tests  made  on  full-length  col- 
FIG.  40.  umns  at  the  Watertown  arsenal : 

(3.) 


Viooz? 


§  5  ARCHITECTURAL  ENGINEERING.  55 

in  which  .V  =  ultimate  comprcssivc  strength  per  square  inch 

of  sectional  area  of  the  column ; 
U  =  ultimate  compressive  strength  of  the  material 

per  square  inch  parallel  to  the  grain ; 
L  =  length  of  column  in  inches; 
D  =  length  of  least  side  of  column  in  inches. 

EXAMPLE. — What  safe  load  will  a  white-pine  column,  10  inches  square 
and  20  feet  l°n£.  support,  using  a  factor  of  safety  of  6  ? 

SOLUTION. — The  ultimate  compressive  strength  of  white  pine  parallel 
to  the  grain  is,  according  to  Table  6,  3,000  pounds  per  square  inch. 
Therefore,  by  substituting  in  the  formula,  we  have 

/3,OOOX240\ 

-s  =  3'00°-(  100x10  )  =  2-~8°P°und*' 

the  ultimate  bearing  value  of  the  column  per  square  inch.  As  the  factor 
of  safety  required  is  6,  the  safe  bearing  value  per  square  inch  of  sec- 
tional area  is  2,280  -e-  6  =  380  pounds.  The  area  of  the  column  being 
100  square  inches,  the  safe  load  is  100  X  380  =  38,000  pounds.  Ans. 

The  column  formulas  in  general  use  do  not  give  a  direct 
method  of  calculating  the  dimensions  of  a  column  to  carry  a 
given  load.  The  usual  method  of  procedure,  when  it  is 
required  to  find  the  dimensions  of  a  column  of  a  given  form 
which  will  safely  support  a  given  load,  is  to  assume  a  value 
for  the  size  of  the  column,  substitute  this  value  in  the  for- 
mula, together  with  the  other  values  given,  and  solve  for  the 
compressive  stress  S.  If  the  assumed  size  gives  a  value  of 
S  that  is  satisfactory  for  the  given  conditions,  it  is  correct ; 
if,  however,  the  resulting  value  of  .S"is  too  great,  the  assumed 
size  is  too  small  and  it  will  be  necessary  to  assume  a  larger 
size  and  try  again ;  if,  on  the  contrary,  the  value  of  .S"  is  less 
than  the  allowable  stress,  a  smaller  size  of  column  may  be 
assumed  and  a  new  value  of  J>  obtained.  After  a  few  trials 
a  size  will  be  found  which  gives  a  satisfactory  unit  stress  for 
the  given  conditions. 

71.  Importance  of  the  Method  of  Securing  the 
Ends  of  Columns. — Materials  in  compression  develop  more 
strength  if  the  bearing  surfaces  are  true  and  level,  for  the 
tendency  is  then  for  all  the  material  in  the  piece,  or  member, 
to  resist  compression  equally,  and  not  crush  in  one  place, 


50  ARCHITECTURAL  ENGINEERING.  §  5 

before  the  balance  of  the  material  can  be  brought  under 
compression,  as  would  be  the  case  if  the  bearing  surfaces 
were  uneven  and  rough. 

The  manner  of  securing  the  ends  of  columns  has,  also,  an 
appreciable  effect  upon  their  strength.  Columns  fixed  so 
firmly  at  the  ends  as  to  be  liable  to  fail  in  the  body  before 
rupturing  their  end  connections,  develop  greater  strength 
than  columns  connected  by  means  of  pins  through, the  ends. 
Columns  with  square  ends  develop  less  ultimate  strength 
than  if  the  ends  are  firmly  fixed,  but  greater  than  if  the  ends 
are  pin-connected,  that  is,  fastened  by  pins  which  permit 
them  to  swing  freely. 

The  above  given  formula  for  wood  columns  and  the  fol- 
lowing formula  for  cast-iron  columns  apply  only  to  those 
columns  having  flat  ends,  the  usual  condition  met  with  in 
building  construction. 

72.  Cast-iron  columns  are  most  frequently  used  in 
buildings  of  moderate  height,  but  have  been,  in  some  cases, 
used  in  buildings  of  sixteen  stories  and  even  more.  The  best 
practice  has,  during  the  last  few  years,  so  uniformly  declared 
in  favor  of  steel  columns  that  the  employment  of  cast  iron 
is  now  generally  confined  to  buildings  of  ordinary  height, 
say,  four  or  five  stories,  or  to  special  cases,  where  advan- 
tages are  to  be  gained  in  the  use,  for  instance,  of  a  number 
of  ornamental  cast  columns. 

It  is  true  that  cast-iron  columns  are  cheaper  per  pound  and 
perhaps  easier  of  erection  than  steel,  though  the  declining 
price  of  structural  steel  is  rapidly  removing  this  considera- 
tion in  favor  of  cast-iron  columns. 

The  uncertain  strength  of  cast  iron  has  compelled  the 
adoption  of  a  very  low  unit  stress,  in  other  words,  a  very 
high  factor  of  safety.  The  uniform  strength  of  structural 
steel  is,  on  the  other  hand,  so  well  understood  that  cast  iron 
is,  for  columns,  and  especially  girders,  falling  into  disuse. 
Considerations  of  economy  may,  however,  in  some  cases, 
still  justify  its  employment. 

One  disadvantage  in  the  use  of  cast-iron  columns  is  that 


§  5  ARCHITECTURAL  ENGINEERING.  57 

when  fracture  occurs,  it  comes  without  warning.  In  high 
buildings,  erected  entirely  upon  cast-iron  columns,  the 
danger  from  wind  pressure  is  very  much  increased  on 
account  of  lack  of  stiffness  in  the  joints  of  the  connections 
at  the  several  floors.  In  fact,  buildings  have  been  blown  10 
inches  out  of  plumb,  owing  to  this  lack  of  rigidity  in  the 
connections. 

73.  Formula  for  Cast-Iron  Columns. — The  strength 
of  a  cast-iron  column  with  square  ends  may  be  calculated 
by  the  following  rule : 

Rule. —  To  find  the  ultimate  strength  per  square  inch  of 
sectional  area  of  a  cast-iron  column  tcith  square  ends,  divide 
the  ultimate  compressive  strength  per  square  inch  of  the 
material  composing  the  column  by  1  plus  the  quotient 
obtained  by  dividing  the  square  of  the  length  of  the  column 
in  inches,  by  3, 600  times  the  square  of  the  radius  of  gyration 
of  the  section  of  the  column. 

This  rule  is  expressed  by  the  formula 


in  which  5  =  ultimate  strength  per  square  inch  of  cross- 
section  ; 

U  =  ultimate  compressive  strength  per  square  inch 
of  the  material   composing  the   column   (for 
cast  iron  U  may  be  taken  as  81,000); 
L  =  length  of  column  in  inches ; 
A72  =  square  of  the  radius  of  gyration. 

The  term  radius  of  gyration,  which  will  be  more  fully 
explained  in  a  later  article,  is  a  mathematical  expression 
much  used  in  calculating  the  strength  of  columns.  Table  !) 
is  a  collection  of  formulas  for  rinding  the  value  of  R*  (the 
square  of  the  radius  of  gyration)  for  the  forms  of  cross- 
section  most  often  used  for  cast-iron  columns. 


58 


ARCHITECTURAL  ENGINEERING. 


TAIJI^E  9. 

TAHL.E  OF  THE  SQUARE  OF  THE  LEAST  RADIUS  OF 

GYRATION  FOR  THE  DIFFERENT  SECTIONS 

OF  CAST-IRON   COLUMNS. 


Square. 


K*  = 


(5.) 


Rectangle.   \. H. 


Solid 
Circular. 

Hollow 
Square. 


12 


(6.) 
(7.) 

(8.) 


Y-A 


Hollow 
Circular. 


EXAMPLE.— What  is  the  square  of  the  radius  of  gyration  of  a  hollow 
circular  column  of  12  inches  outside  diameter,  the  thickness  of  the 
metal  being  1  inch  ? 

SOLUTION. — The  formula  for  obtaining  the  square  of  the  radius  of 
gyration  for  a  hollow  cylindrical  cast-iron  column  is,  according  toTable  9, 

__  n*  +  A* 
~16  — ' 
Substituting  the  given  values,  we  have 

244 


— 16 —      -  -jg-  =  15.2.      Ans. 

EXAMPLE.— Find  the  proper  working  load  for  a  10-inch  square  hollow 
cast-iron  column,  20  feet  long,  using  a  factor  of  safety  of  6,  the  thick- 
ness of  the  metal  being  1  inch. 

SOLUTION.— The  ultimate  compressive  strength  U  of  cast-iron  per 
luare  inch,  according  to  Table  6,  is  81,000  pounds,  the  length  L  is 
20  X  12  =  240  in.  ;  and  A"  for  a  hollow  square  rectangular  column  is 
according  to  Table  9, 

10"  + 8s 


12 


ARCHITECTURAL    ENGINEERING. 


Substituting  these  values  in  formula  -A,  \\e  have 

81.000  _  81.000  _ 

:  1T17" 


5  = 


3,600 A'-  '  3,600X13.67 

the  breaking  strength  of  the  column  in  pounds  per  square  inch  of 
section.  AVith  a  factor  of  safety  of  6,  the  safe  bearing  value  of  the 
column  is  37,32?  -=-  6  =  6,221  pounds  per  square  inch.  The  net  area  of 
the  section  of  the  column  is  10-  —  8'-  =  100  —  64  =  36  square  inches. 
The  entire  load  that  it  will  support  with  safety  is,  therefore,  36  X  6,221 
=  223,956  pounds.  Ans. 


DESIGN  OF  CAST-IRON  COLUMNS. 

74.     Fig.   41   shows   a  design   for  a  circular  cast-iron 
column.     B  shows  the  elevation  for  the  cap  and  brackets 


Wood  Column. 


Cant  Iron  Column. 


Plan  of  Base. 


FIG.  41. 


supporting-  steel  floorbeams.  Attention  should  be  paid  to 
the  design  of  the  bracket  a  and  the  web  made  as  shown  at 
B,  care  being  taken  that  it  is  brought  to  the  edge  of  the 
plate  ;«,  upon  which  the  steel  beams  rest.  Otherwise,  if 
the  beam  takes  a  bearing  on  the  edge  of  the  plate  ///  as 


60  ARCHITECTURAL  ENGINEERING.  §  5 

shown  at  C,  the  tendency  will  be  to  fracture  the  edge  of  the 
bracket.  The  web  of  the  bracket  should,  if  possible,  form  an 
angle  of  60°  with  the  horizontal  plate  ;//,  as  shown  in  Fig.  41. 

The  bolt  holes /should  be  always  drilled  either  in  the 
casting  or  in  the  steel  beams  after  these  beams  are  in  place, 
because  if  the  holes  were  cored  in  the  casting  and  the  holes 
punched  in  the  beams  at  the  mill,  it  would  more  than  likely 
be  found  in  the  course  of  erection  at  the  building  that  the 
beams  were  supported  entirely  by  the  shear  of  the  bolts  and 
would  not  bear  upon  the  supporting  brackets.  It  is  well  to 
have  at  least  ^-inch  fillets,  as  shown,  but  larger  if  possible, 
in  all  the  corners  of  the  casting.  It  is  good  practice  to 
thicken  up  the  metal  .in  the  column,  where  the  brackets  are 
cast  upon  it,  as  shown  at  b. 

In  forming  the  bolt  holes/,  it  must  be  remembered  that 
the  bolt  should  fit  the  hole  as  closely  as  possible ;  that  it 
should,  in  fact,  be  a  machine  fit.  It  is  well,  indeed,  to  drill 
the  holes  both  in  the  beams  and  the  cast-iron  flanges,  to 
insure  a  true  and  accurate  bolt  hole,  so  that  there  may  be  a 
minimum  amount  of  play  in  the  connection  and  that  it  may 
be  as  rigid  as  possible. 

In  designing  the  base  for  this  column,  it  is  well  to  place 
the  ribs  strengthening  the  web  in  such  position  that  they 
may  be  most  effective,  which,  in  this  case,  is  on  the  diagonals, 
for,  if  placed  parallel  to  the  sides,  the  corners  will  have  a 
tendency  to  break  off,  and  part  of  the  bearing  surface  at  the 
base  of  the  column  may  prove  ineffective. 

Designs  for  the  base  and  cap  of  cast-iron  columns  are  as 
numerous  as  various  conditions  may  require  them.  It  would 
be  almost  impossible  to  give  examples  of  the  different  con- 
nections demanded  in  actual  building  practice.  The  fore- 
going remarks  will,  if  kept  in  mind,  apply  to  every  case,  and, 
if  followed,  insure  good  design  as  well  as  secure  construction. 

75.  Inspection. — In  examining  castings  used  in  build- 
ing construction,  to  ascertain  their  quality  and  soundness, 
several  points  are  to  be  considered.  The  edges  should  be 
struck  with  a  light  hammer.  If  the  blow  makes  a  slight 


§5  ARCHITECTURAL  ENGINEERING.  01 

impression,  the  iron  is  probably  of  good  quality,  providing 
it  be  uniform  throughout.  If  fragments  fly  off  and  no  sen- 
sible indentation  be  made,  the  iron  is  hard  and  brittle.  Air 
bubbles  and  blowholes  are  a  common  and  dangerous  source 
of  weakness.  They  should  be  searched  for  by  tapping  the 
surface  of  the  casting  all  over  with  a  hammer.  Bubbles,  or 
flaws,  filled  in  with  sand  from  the  mold,  orptirposely  stopped 
with  loam,  cause  a  dullness  in  the  sound,  leading  to  their 
detection.  The  metal  of  a  casting  should  be  free  from 
bubbles,  core  nails,  or  flaws  of  any  kind.  The  exterior  surface 
should  be  smooth  and  clean,  and  the  edges  of  the  casting 
should  be  sharp  and  perfect.  An  uneven  or  wavy  surface 
indicates  unequal  shrinkage,  caused  by  want  of  uniformity 
in  the  texture  of  the  iron. 

The  surface  of  a  fracture,  examined  before  becoming  rusty, 
should  present  a  fine-grained  texture,  of  a  uniform  bluish- 
gray  color  and  high  metallic  luster. 

In  inspecting  cast-iron  columns,  care  should  be  taken  to 
see  that  they  are  straight  and  cored  directly  through  the 
center,  and  that  the  metal  is  of  the  same  thickness  through- 
out. It  is  not  unusual  to  find,  in  molding  cast-iron  columns, 
that  the  core  has  not  been  placed  in  the  mold  in  a  central 
position,  or  that,  having  been  insecurely  fastened,  it  has 
floated  over  to  one  side.  Hence,  the  column  which  should 
have  been,  say,  f  of  an  inch  in  thickness  throughout,  maybe 
1^  inches  on  one  side  and  ^  of  an  inch  on  the  other.  The 
base  and  the  cap  of  a  cast-iron  column  should  be  turned 
accurately,  being  true  and  perpendicular  to  the  center  line 
of  the  column. 

76.  Banger  from  Fire. — One  of  the  great  objections 
to  the  use  of  cast-iron  columns  is  that  they  are  liable  to  be 
broken  by  sudden  contraction,  due  to  water  being  played  upon 
them  in  case  of  fire. 

EXAMPLES   FOR  PRACTICE. 

1.  A  yellow-pine  column  20  feet  long  is  required  to  sustain  a  load 
of  100,000  pounds;  provided  a  factor  of  safety  of  5  is  used,  what  must 
be  the  size  of  this  column  ?  Ans.  12  in.  X  12  in. 


$•>  ARCHITECTURAL  ENGINEERING.  §  5 

2.  What  will  be  the  allowable  load  upon  a  4"  X  10"  spruce  column, 
12  feet  long,  the  factor  of  safety  being  6  ?  Ans.    12,800  Ib. 

3.  It  is  required  that  a  round  yellow-pine  column  shall  carry  173,000 
pounds.     What  must  be  the  diameter  of  the  column,  the  safe  unit  com- 
pressive  stress  upon  the  material  being  1,000  pounds  ?  Ans.  15  in. 

4.  The  section  of  a  hollow  cast-iron  column  is  16  inches  square, 
outside  measurement;  the  thickness  of  the  material  in  the  column  is 
1  inch ;  what  is  the  radius  of  gyration  of  this  column  ?  Ans.  6.13. 

5.  What  will  be  the  breaking  load  of  a  cast-iron  column  20  feet  long, 
12  inches  in  diameter  outside,  made  of  1-inch  metal  ?      Ans.   1,372,200  Ib. 

6.  The  thickness  of  the  metal  in  a  cast-iron  column  is  f  inch ;  if  a 
factor  of  safety  of  6  is  used,  and  the  length  of  the  column  is  18  feet, 
what  must  be  the  outside  diameter  of  the  column  to  support  a  load  of 
133,000  pounds  ?  Ans.   10  in. 

7.  Find  the  ultimate  crushing  strength  of  a  10-inch  square,  outside 
measurement,  cast-iron  column;  the  thickness  of  the  metal  is  1  inch, 
and  the  length  of  the  column  is  20  feet.  Ans.  1,343,700  Ib. 


BEAMS. 


DEFINITIONS. 

77.  Any  harvesting  upon  supports,  and  liable  to  trans- 
verse stresses,  is  called  a  beam. 

A  simple  beam  is  a  beam  resting  upon  two  supports  very 
near  its  ends. 

A  cantilever  is  a  beam  resting  upon  one  support  in  its 
middle,  or  which  has  one  end  fixed  (as  in  a  wall)  and  the 
other  end  free. 

A  fixed  beam  is  one  which  has  both  ends  firmly  secured 
(as  a  plate  riveted  to  its  supports  at  both  ends). 

A  continuous  beam  is  one  which  rests  upon  more  than 
two  supports. 

The  span  of  a  simple  beam  is  the  distance  between  its 
supports. 

REACTIONS. 

78.  Since  one  condition  of  equilibrium  requires  that  the 
sum  of  all  the  forces  acting  on  a  body  in  one  .direction  must 
be  balanced  by  an  equal  set  of  forces  acting  in  the  opposite 


§5  ARCHITECTURAL  ENGINEERING.  63 

direction,  it  follows  that,  in  order  that  any  body  may  be 
kept  from  falling,  there  must  be  an  upward  pressure,  or 
thrust,  against  it,  just  equal  to  the  downward  pressure  due 
to  its  weight;  this  upward  thrust  is  called  a  reaction. 

In  accordance  with  this  principle,  it  is  evident  that  the 
simple  beam   shown  in   Fig.  ±2   is  supported  by  the  sum  of 


the  upward  pressures  exerted  on  it  by  the  two  brick  piers  on 
which  it  rests;  also  that  this  sum  is  equal  to  the  weight  of 
the  beam  plus  the  weight  of  any  load  it  may  carry.  This  is 
expressed  by  the  statement:  The  sum  of  t lie  reactions  at  tlie 
supports  of  any  beam  is  equal  to  the  sum  of  the  loads. 

7J).  Relation  Uetween  the  Reactions. — If  the  load  on 
a  simple  beam  is  either  uniformly  distributed,  applied  at  the 
center  of  the  span,  or  symmetrically  placed  on  each  side  of  the 
center  of  the  span,  the  reaction  at  each  support  is  equal  to 
one-half  of  the  total  load.  When,  however,  the  loads  are  not 
symmetrically  placed,  the  reactions  are  unequal  and  must 
be  determined  before  the  first  step  towards  obtaining  the 
strength  of  the  beam  can  be  taken.  To  determine  the  reac- 
tions at  the  points  of  support  of  a  beam  loaded  with  a 
number  of  loads,  irregularly  placed,  we  apply  the  principle 
of  moments,  as  shown  in  the  following  illustrative  examples: 

8O.  Two  men  a  and  d,  15  feet  apart,  carry  a  50-pound 
weight  between  them  on  a  plank,  as  shown  in  Fig.  43.  What 
part  of  the  load  does  each  man  carry? 

If  the  load  had  been  placed  midway  between  them,  it  is 
quite  evident  that  each  man  would  have  half  the  weight  of 
the  plank'and  load  to  support.  But,  since  the  load  is  moved 
towards  a  until  within  5  feet  of  him,  he  must  support  a 
greater  proportion  of  the  load  than  b.  If  b  raises  his  end  of 


64 


ARCHITECTURAL  ENGINEERING. 


the  plank,  as  shown  in  dotted  lines,  it  is  evident  that  a  simply 
acts  as  a  hinge  while  b  raises  the  weight  with  a  lever  15  feet 
long.  The  weight  of  50  pounds  acts  down  with  a  leverage 
of  5  feet ;  its  moment  about  a  as  a  center  is,  therefore,  50x5 
=  250  foot-pounds.  That  there  may  be  equilibrium  of 


moments,  it  is  evident  that  the  man  at  b  must  exert  an 
upward  force  whose  moment  with  a  lever  arm  of  15  feet 
equals  250  foot-pounds;  that  is,  he  must  exert  a  force  of 
250 -=-15  =  16f  pounds  to  support  his  share  of  the  weight. 
Since  the  sum  of  the  reactions  must  equal  the  sum  of  the 
loads,  it  follows  that  if  b  supports  16§  pounds,  a  must  sup- 
port the  difference  between  the  load  of  50  pounds  and  16§ 
pounds,  or  33£  pounds. 

81.  Fig.  44  shows  the  men  a  and  b  supporting  three 
loads  of  50,  40,  and  80  pounds,  respectively.  It  is  desired 
to  estimate  the  force  that  each  must  exert  to  sustain  the 
weights,  leaving  the  weight  of  the  plank  out  of  the  question. 
Assuming  the  center  of  moments  at  a,  find  the  resultant 
moment  of  all  the  weights  about  this  point,  as  follows  : 

50  X  3  =  150  ft.-lb. 
40  X  8  =  320  ft.-lb. 
80x12  -960  ft.-lb. 

Total,     1,4.3  0  ft.-lb. 


ARCHITECTURAL  ENGINEERING. 


05 


This  is  the  moment  of  all  the  loads  upon  the  beam  about 
the  point  a  as  a  center.     Hence,  the  force  that  b  must  exert,  in 


«                                                     IS  ft 

1   ft 

•  ~3ft  —  ^ 

1 

nw\         °  \ 

* 

n 

-  .^ 

FIG.  44. 


order  to  produce  equilibrium,  is  1,4.30 -=-15  =  !)5^  pounds. 
The  part  of  the  load  which  a  supports  is  the  difference  between 
the  total  load,  50  +  40  +  80  =  170  pounds,  and  the  part  of 
the  load  supported  by  d,  that  is,  170  —  95^  =  74f|  pounds. 


/"A 


-V" 
\ 


FIG.  45. 


82.     Take  a  more  practical  example.     In  Fig".  45  let  it 
be  required  to  find  the  reactions  Rl  and  A'2.      (In  all  the 


CO 


ARCHITECTURAL  ENGINEERING. 


subjoined  problems,  Rt  and  Rn  represent  the  reactions.)  The 
center  of  moments  may  be  taken  at  either  Rt  or  Rn.  Taking 
Rt  as  the  center  in  this  case,  construct  a  diagram  as  in  Fig. 


/ 

/ 


/ 

-    .   O^  ft    . 

> 

* 

* 

*—•                                —  10  ft.  — 

>• 

"  5ft--> 

> 

1 
1 

FIG.  46. 

46.  The  three  loads  are  forces  acting  in  a  downward  direc- 
tion ;  the  sum  of  their  moments  with  respect  to  the  assumed 
center  may  be  computed  as  follows: 

8,000x  5  =  4  0,0  0  0  ft.-lb. 
6,000x19  =  1  1  4,0  0  0  ft.-lb. 
2,000x27  =  5  4,0  00  ft.-lb. 

Total,  2  0  8,0  0  0  ft.-lb. 

The  magnitude  of  the  reaction  7?,  acting  in  an  upward 
direction  with  a  lever  arm  of  30  feet  is,  therefore,  208,000 
-^-30  =  6,933$  pounds.  The  sum  of  all  the  loads  is  2,000 
+  6,000  +  8,000  =  16,000  pounds.  Then  the  reaction  at  R^ 
is  16,000-6,933$  =  9,066|  pounds. 

EXAMPLE  1.— What  is  the  reaction  at  R*  in  Fig.  47  ? 


-                    —  xvji.  •- 

9> 

tcr  of  Gravity 
'niforin  Loud. 

4J 

»          '  '  >  3  OOO  I  b.  pe  f  r  u  n  n  i  n  <i  ft. 

*r*l                                                                             1  r" 
\                    \  '*"       "          '-*                      \ofl 

„ 

J 

SOLUTION.— In  computing  the  moment  due  to  a  uniform  or  evenly 
distributed  load,  as  at  a,  the  lever  arm  is  always  considered  as  the  dis- 
tance from  the  center  of  moments  to  the  center  of  gravity  of  the  load. 


§  5  ARCHITECTURAL  ENGINEERING.  67 

The  amount  of  the  uniform  load  a  is  3,000  X  10  =  30,000  pounds,  and 
the  distance  of  its  center  of  gravity  from  A\  is  13  feet.  The  moments  of 
the  loads  upon  this  beam  about  7^  are  as  follows: 

30,000  X  13  =  B  9  0,0  0  0  ft.-lb. 
4,000  X  4  =  1  6,00  0  ft.-lb. 
9,000  X  20  =  1  8  0,0  0  0  ft.-lb. 


Total,         586,000  ft.-lb. 

This  is  the  sum  of  the  moments  of  all  the  loads  about  7\}  as  a  center. 
The  leverage  of  the  reaction  A'.2  is  30  feet.  Hence,  the  reaction  at 
A%  is  586,000  -H  30  =  19,533i  pounds.  Ans. 


*  — 

•*  2% 

-  —  —  is  ft. 

•3  ft^2OOOO  Ib. 

—  1               \3000  Iff. 
2000  /ft|              | 

.5000  Ib. 

' 

*1OOO  Ib.  per  running  ft 

*, 

RI 

FIG.  48. 

EXAMPLE  2. — A  beam  is  loaded  as  shown  in  Fig.  48.  Compute  the 
reactions  A'i  and  AV 

SOLUTION. — Consider  A*]  as  the  center  of  moments.  Then  the  moments 
of  the  loads  about  A'i  are: 

20,000  X    3  =      60,000  ft.-lb. 

2,000X18  =      36,000  ft.-lb. 

3,000X22  =      66,000  ft.-lb. 

5,000X^6  =  1  80,000  ft.-lb. 

1,000  X  6  X  33  =  1  9  8,0  0  0  ft.-lb. 

Total,  540,000  ft.-lb. 

This  divided  by  30,  the  length  of  the  lever  arm  of  the  reaction  R^ 
gives  18,000  pounds,  the  reaction  at  A*2.  The  sum  of  the  loads  is  20,000 
4-  2,000  +  3,000  +  5,000  +  6,000  =  36,000  pounds;  and  36,000  -  18,000 
=  18,000  pounds,  the  other  reaction  A*,.  Ans. 

EXAMPLE  3. — Compute  the  reactions  at  the  supports  R^  and  R?  in  a 
beam  loaded  as  shown  in  Fig.  49. 

SOLUTION. — Again,  letting  A'i  be  the  center  of  moments,  the  moments 
of  the  loads  are : 

5,000X10  =  50,000  ft.-lb. 
10,000  X  20  =  2  0  0,0  0  0  ft.-lb. 
30,000  X  40  =  1,2  0  0,0  0  0  ft.-lb. 

Total,         1,4  5  0,0  0  0  ft.-lb. 
1-17 


G8  ARCHITECTURAL  ENGINEERING.  §  5 

Now,  1,450,000  -T-  30,  the  distance  between  the  supports  =  48,333£ 
pounds,  the  required  reaction  at  /?2.  The  sum  of  the  loads  is  5,000 
+  10,000  +  30,000  =  45,000  pounds ;  therefore,  the  reaction  R*  is  greater 
than  the  sum  of  the  loads.  This  shows  that  the  force  at  Rl  must  act  in 
a  downward  direction  in  order  that  the  sum  of  the  downward  forces 
may  equal  the  upward  force  at  Rt.  Since  this  is  opposite  to  the  usual 
direction,  the  reaction  at  Rl  is  called  negative  or  minus.  In  other  words, 
instead  of  an  upward  reaction  at  Rt,  there  must  be  a  downward  force  at 
this  point,  or  the  beam  will,  as  shown  by  the  dotted  lines,  rotate  around 


a 

•2 — 40  /<r- 


w 

I*—  —  §— 20ft. — 


f          

-30ft. —  — V—    —10ft. < 


FIG.  49. 

the  support  R*.  The  magnitude  of  this  downward  force  is  the  differ- 
ence between  the  up\vard  reaction  at  A'2  and  the  sum  of  the  downward 
pressures  due  to  the  loads;  that  is,  48,333£  —  45,000  =  3,3331  pounds. 

We  will  now  compute  the  reaction  at  RI  by  taking,  the  center  of 
moments  at  /?2,  and  applying  the  rule  in  Art.  35  to  find  the  magnitude 
and  direction  of  action  of  the  force  at  R?  whose  moment  is  the  resultant 
of  the  moments  of  the  loads  on  the  beam.  The  load  of  30,000  pounds 
tends  to  produce  right-hand  rotation  around  the  center  R* ;  hence,  its 
moment,  30,000  X  10  =  300,000  foot-pounds,  is  positive.  The  10,000- 
pound  load  is  10  feet  to  the  left  of  /?2,  and  its  tendency  is  to  produce 
left-hand  rotation  about  Rz ;  consequently,  its  moment  is  negative  and 
equal  to  10,000  X  10  =  100,000  foot-pounds.  We  find,  in  a  similar 
manner,  the  moment  of  the  5,000-pound  load  to  be  negative  and  equal  to 
5,000  X  20  =  100,000  foot-pounds.  These  results  may  be  collected  thus : 
Positive  moment: 

30,000X10=          1     .     .     +300,000  ft. -Ib. 

Negative  moments: 

10,000  X  10  =  1  0  0,0  0  0  ft.-lb. 
5,000  X  20  =  1  0  0,0  0  0  ft.-lb. 

-200,000  ft.-lb. 
Algebraic  sum,     +  1  0  0,0  0  0  ft.-lb., 


§5  ARCHITECTURAL  ENGINEERING.  C9 

the  resultant  of  the  moments  of  the  three  loads.  Since  the  positive 
moment  is  greater  than  the  sum  of  the  negative  moments,  we  see  that 
to  produce  equilibrium  the  force  at  A\  must  tend  to  produce  left-hand 
rotation ;  that  is,  it  must  act  downwards ;  its  lever  arm  being  30  feet 
long,  its  magnitude  must  be  100,000 -v- 30  =  3,333^  pounds,  the  same 
result  as  was  obtained  before. 


EXAMPLES   FOR   PRACTICE. 

1.  The  span  of  a  simple  beam  is  25  feet;   at  distances  of  9  feet,  16 
feet,  and  18  feet  from  the  left-hand  end  are  placed  concentrated  loads 
of  8,000,  4,000,  and  16,000  pounds,  respectively.     What  is  the  magnitude 
of  the  left  reaction  ?  Ans.   11,040  Ib. 

2.  The  two  reactions  supporting  a  beam  are  2,500  and  3,000  pounds ; 
what  is  the  single  concentrated  load  necessary  to  produce  these  reac- 
tions ?  Ans.   5,500  Ib. 

3.  The  length  of  a  beam  is  30  feet,  and  it  overhangs  the  right-hand 
support  6  feet.     From  the  overhanging  end  there  is  a  weight  of  6,000 
pounds;   10  feet,  12  feet,  and  18  feet  from  the  left-hand  support  are 
loads  of  8,000,  6,200,  and  7,800  pounds,  respectively.     What  is  the  mag- 
nitude of  the  right-hand  reaction  ?  Ans.   19,783i  Ib. 

4.  If  for  a  distance  of  10  feet  from  the  left-hand  end  of  a  beam  is 
distributed  a  load  of  1,000  pounds  per  running  foot,  and  at  the  center 
of  the  beam  is  located  the  concentrated  load  of  16,500  pounds,  what  is 
the  magnitude  of  the  left-hand  reaction,  providing  the  beam  is  sup- 
ported at  both  ends  and  is  30  feet  long  ?  Ans.  16,583^  Ib. 


STRESSES    IX    BEAMS. 

83.  We  have  seen  that  a  beam  is  a  body  acted  on  by 
various  external  forces  so  related  as  to  be  in  a  condition  of 
equilibrium ;  so  far,  however,  we  have  not  considered  the 
effect  of  these  forces  on  the  beam  itself. 

In  a  body  subjected  to  a  direct  pull  or  thrust,  as  a  rope  or 
a  column,  the  external  forces  are  directly  opposed  to  each 
other,  and  the  resultant  stresses  in  all  sections  are  of  the 
same  kind,  tension  or  compression.  In  a  beam,  however, 
the  external  forces,  while  they  generally  act  in  parallel 
lines,  are  not  directly  opposed  to  each  other,  and  it  is  the 
function  of  the  beam  to  transfer  these  forces  from  one 


?0  ARCHITECTURAL  ENGINEERING.  §  5 

line  of  action  to  another.  Take,  for  example,  the  case  of  a 
weight  suspended  from  a  pin  driven  in  a  wall,  as  shown  in 
Fig.  50.  The  downward  force  of  20  pounds 
due  to  the  action  of  the  weight  is  balanced  by 
the  upward  pressure  or  reaction  of  the  pin  on 
the  rope;  the  rope  is  thus  subjected  to  the 
action  of  two  directly  opposing  forces,  the 
result  being  a  tensile  stress  which  is  the  same 
for  each  section  of  the  rope  between  the  weight 
and  pin.  The  pin  acts  as  a  cantilever  beam 
which  transfers  horizontally  to  the  wall,  where 
it  is  balanced  by  an  equal,  upward  pressure  or 
reaction,  the  downward  pressure  due  to  the 
pull  of  the  rope.  The  pin  is  thus  subjected  to 
two  opposing  forces  which,  however,  act  in 
different  lines;  these  forces  produce  a  set  of 
[20  ib  opposing  forces,  or  stresses,  in  the  pin  itself, 
which  are  different  in  kind  for  different  parts 
of  the  pin,  and  vary  in  magnitude  for  each 
section  between  the  rope  and  the  wall. 


SHEAR. 

84.  By  an  inspection  of  Fig.  50,  we  see  that  if  we   pass 
a  vertical  plane  through  any  point  between  the  rope  and 
wall,  the  part  of  the  pin  between  this  plane  and  the  rope 
will  be  acted  on  by  a  downward  force  due  to  the  pull  of  the 
rope,  while  the  other  is  subjected  to  an  equal  upward  force 
due  to  the  reaction  of  the  wall ;    the  action  of  these   two 
forces  tends  to  slide  the  two  parts  of  the  pin  past  each  other, 
along  the  section  formed  by  the  cutting  plane.     The  pin  is 
thus  subject  to  a  stress  which,  from  its  similarity  to  a  shear- 
ing action,  is  called  shear. 

85.  Shear    in   a  Simple  Beam. — Consider    now    the 
simple  beam  shown  in  Fig.  51.     Since  the  loads  are  sym- 
metrically applied,  each  reaction  is  equal  to  40  pounds,  one- 
half  of  the  total  load  on  the  beam.      Beginning  at  the  left 
reaction  Rt,  there  is  an  upward  force  of  40  pounds  acting  on 


§  5  ARCHITECTURAL  ENGINEERING.  71 

the  beam ;  since  the  forces  are  in  equilibrium,  this  upward 
force  is  balanced  by  an  equal  downward  force,  which  is  the 
vertical  resultant  of  the  loads  and  the  reaction  R^.  Consid- 
ering, therefore,  any  section  of  the  beam  between  R1  and 
the  point  of  application  of  the  load  ;/,  we  see  that  the  part  of 
the  beam  at  the  left  of  this  section  is  subjected  to  an  upward 
thrust  of  40  pounds,  while  the  part  at  the  right  is  subjected 
to  an  equal  downward  thrust ;  the  result  is  a  shearing  stress 
on  this  section,  whose  magnitude  is  equal  to  the  reaction  R^ 


»  13 

*"l       .»        Si 


FIG.  51. 


When  the  point  of  application  of  ;/  is  reached,  the  effect 
of  the  upward  force  Rt  is  partly  balanced  by  the  downward 
force  of  10  pounds  due  to  the  load  n  ;  considering,  therefore, 
any  section  of  the  beam,  as  a  b,  between  the  points  of  appli- 
cation of  the  loads  n  and  m,  we  see  that  the  part  of  the  beam 
at  the  left  is  acted  on  by  the  vertical  resultant  of  the  reac- 
tion Rt  and  the  load  n,  that  is,  by  an  upward  force  of  40 
—  10  =  30  pounds,  while  the  part  at  the  right  is  acted  on  by 
an  equal  downward  force,  the  vertical  resultant  of  the 
remaining  loads  and  the  reaction  R^.  Any  section  between 
the  points  of  application  of  n  and  m  is,  therefore,  subject  to 
a  shearing  stress  equal  to  the  difference  between  the  reac- 
tion R1  and  the  load  n,  that  is,  to  40  —  10  =  30  pounds. 
In  the  same  way,  it  follows  that  the  shearing  stress  for  any 
section  between  m  and  o  is  40  —  (10  +  15)  =  15  pounds.  For 
any  section,  as  c  d,  between  the  points  of  application  of  o 
and  /,  the  shearing  stress  is  40  —  (10  +  15  +  15)  =  0;  in 
other  words,  on  each  side  of  this  section  the  downward 
forces  and  the  reactions  are  equal,  and  their  resultant  is 
zero;  it  is,  therefore,  a  section  in  which  there  is  no  shear. 


72  ARCHITECTURAL  ENGINEERING.  §  5 

For  convenience,  it  is  customary  to  call  the  reactions,  or 
forces,  acting  in  an  upward  direction,  positive,  and  the 
loads,  or  downward  forces,  negative;  since  the  difference 
between  the  sums  of  the  positive  and  negative  numbers  rep- 
resenting a  given  set  of  values  is  called  their  algebraic  sum, 
it  follows  that  the  shear  for  any  section  of  a  beam  is  equal  to 
the  algebraic  sum  of  either  reaction  and  the  loads  between  this 
reaction  and  the  given  section. 

In  nearly  all  cases  the  external  forces — loads  and  reac- 
tions— act  on  a  beam  along  vertical  lines ;  the  shearing  stress 
just  considered  being  the  resultant  of  these  forces  along  a 
section  formed  by  an  imaginary  vertical  cutting  plane,  is 
often  called  the  A-ertical  shear. 

86.  Maximum  Shear. — From  what  has  been  said,  it  is 
evident  that  the  shear  in  any  simple  beam  is  always  greatest 
between  the  reactions  and  the  nearest  loads,  and  that  in  any 
case  the  maximum  shear  is  equal  to  the  greater  reaction. 

87.  Positive  and  Negative  Shear. — If  we  take  a  sec- 
tion of  the  beam  near  the  left  reaction  and  consider  the 
forces  acting  on  the  part  of  the  beam  at  the  left,  we  see  that 
their  resultant  is  positive;  the  shear  at  this  section  is,  there- 
fore, called  positive  shear.     If,  however,  we  take  a  section 
near  the  right  reaction,  the  resultant  of  the  forces  at  the  left 
is  found  to  be  negative,   and  in  consequence  the  shear  is 
called  negative.     It  is  also  evident  that  there  is  a  section 
between  the  two,  where  the  resultant  of  the  forces  changes 
from  positive  to  negative ;  at  such  a  section  the  shear  is  said 
to  change  sign. 

EXAMPLE  1. — (a)  What  is  the  maximum  shear  on  the  beam  shown  in 
Fig.  52  ?  (b)  What  is  the  shear  at  a  point  9  feet  from  the  right  support  ? 
and  (c)  what  is  the  shear  at  a  point  18  feet  from  the  right  support  ? 

SOLUTION.—  (a)  First  estimate  the  reactions  as  follows:  Taking  the 
center  of  moments  at  the  left  support,  the  moments  of  the  loads  are: 

2,000  X    3  =         6,0  0  0  ft.-lb. 

6,000X11  =      66,000  ft.-lb. 

8,000  X  25  =  2  0  0,0  0  0  ft.-lb. 

Total,       2  7  2,0  0  0  ft.-lb. 


ARCHITECTURAL    ENGINEERING. 


73 


272,000  -H  30  =  9,066|  pounds,  the  reaction  at  AY  The  sum  of  the  loads 
equals  2,000  +  6,000  +  8,000  =  16,000  pounds;  16,000 -9,066f  =  6,9331 
pounds,  the  reaction  at  AY  The  maximum  shear  is  therefore  9,066| 
pounds.  Ans.  (b)  As  the  reaction  A'2  at  the  right  support  is  equal  to 
9,066|  pounds,  and  as  there  is  only  the  one  load  c  of  8,000  pounds, 

S  5  * 

«  c  c 

a  o  X 

O> 


e 

»                                                           ao 

•O                                                                                o 

—  9ft— 

~  * 

FIG.  52. 

between  A*2  and  a  point  9  feet  away,  the  shear  at  this  point  must  equal 
9,066$- 8,000  =-l,066§  pounds.  Ans.  (c)  The  shear  at  18  feet  from 
the  reaction  A*2  is  also  1,066|  pounds,  because  there  is  no  other  weight 
occurring  between  this  point  and  AY 

EXAMPLE  2. — At  what  point  in  the  beam  loaded  as  shown  in  Fig.  53 
does  the  shear  change  sign  ? 


^3000  Ib.per  runn\ing  ft. 
14.48ft- 

30ft. 


FIG.  53. 

SOLUTION. — Compute  the  reaction  A',  as  follows:  With  the  center  of 
moments  at  A^  the  moments  of  the  loads  are : 

9,000  X  10  =      9  0,0  0  0  ft.-lb. 

4,000  X  26  =  1  0  4,0  0  0  ft.-lb. 

3,000  X  10  X  17  =  5  1  0.0  0  0  ft.-lb. 

Total,          704,000  ft.-lb. 

704, 000 -j- 30  =  23,466|  pounds,  the  reaction  at  A',.  The  first  load  that 
occurs,  working  out  on  the  beam  from  AY  is  c  of  4,000  pounds.  Then, 
23,466f  —  4,000  =  19,466|  pounds.  The  next  load  that  occurs  on  the 
beam  is  the  uniform  load  of  3,000  pounds  per  running  foot.  There 


74  ARCHITECTURAL  ENGINEERING.  §  5 

being  altogether  30,000  pounds  in  this  load,  it  is  evident  that  it  will 
more  than  absorb  the  remaining  amount  of  the  reaction  Ri ;  the  point 
where  the  change  of  sign  occurs  must  consequently  be  somewhere  in 
that  part  of  the  beam  covered  by  the  uniform  load.  The  load  being 
3,000  pounds  per  running  foot,  if  the  remaining  part  of  the  reaction, 
19,466f  pounds,  be  divided  by  the  3,000  pounds,  the  result  will  be  the 
number  of  feet  of  the  uniform  load  required  to  absorb  the  remaining 
part  of  the  reaction,  and  this  will  give  the  distance  of  the  section, 
beyond  which  the  resultant  of  the  forces  at  the  left  becomes  negative, 
from  the  edge  of  the  uniform  load  at  a;  thus,  19,466f -?- 3,000  =  6.48 
feet.  The  distance  from  Rl  to  the  edge  of  the  uniform  load  is  8  feet. 
The  entire  distance  to  the  section  of  change  of  sign  of  the  shear  is, 
therefore,  8  +  6.48  =  14.48  feet  from  RL  Ans. 


EXAMPLES  FOB  PRACTICE. 

1.  The  uniformly  distributed  load  upon  a  beam  supported  at  both 
ends  is  40,000  pounds.     What  is  the  maximum  shear  upon  the  beam  ? 

Ans.  20,000  Ib. 

2.  A  beam  is  loaded  with  three  concentrated  loads:    A  of  2,000 
pounds,  B  of  6,000  pounds,  and  C of  8,000  pounds;  they  are  located  10 
feet,  12  feet,  and  18  feet,  respectively,  from  the  left-hand  end  of  the 
beam,  the  span  of  which  is  40  feet.     What  is  the  shear  between  the 
loads  C  and  Bl          k  Ans.  2,100  Ib. 

3.  The  span  of  a  beam  is  20  feet,  and  there  is  a  uniformly  distributed 
load  on  three-quarters  of  this  distance  from  the  left-hand  support  of 
9,000  pounds.     At  distances  of  8  feet  and  12  feet  from  the  right-hand 
support  are  located  concentrated  loads  of  5, 000  pounds  and  6,000  pounds, 
respectively.     At  what  distance  from  the  left-hand  end  of  the  beam 
does  the  shear  change  sign  ?  Ans.  8  ft. 


BENDING  STRESSES. 

88.  Bending  Moment. — If,  in  a  cantilever  loaded  as 
in  Fig.  54,  we  take  any  point  x  on  the  center  line  a  b,  as  a 
center  of  moments,  and  consider  a  section  made  by  a 
vertical  plane  c  d  through  this  center,  it  is  evident  the 
moment  of  the  force  due  to  the  downward  thrust  of  the  load 
tends  to  turn  the  end  of  the  beam  to  the  right  of  c  d,  around 
the  center  x ;  the  measure  of  this  tendency  is  the  product 
of  the  weight  W  multiplied  by  its  distance  from  c  d,  and, 


§5  ARCHITECTURAL  ENGINEERING.  75 

since  it  is  the  moment  of  a  force  which  tends  to  bend  the 
beam,  it  is  called  the  bending  moment. 

89.  Resisting  Moment. — A  further  inspection  of  Fig. 
54  shows  that  if  the  end  of  the  beam  turns  around  the 
center  .r  until  it  takes  the  position  shown  by  the  dotted 
lines,  the  parts  of  the 
two  surfaces  formed 
by  the  cutting  plane 
c  d  which  are  above 
the  center  ,r,  must  be 
pulled  away  from 
each  other,  while 
those  below  are 
pushed  closer  to-  FIO.  r>j. 

gether.  We  thus  see  that,  if  we  consider  a  vertical  section 
through  any  point  on  the  center  line  a  b,  between  the  load 
and  the  point  of  support,  the  tendency  of  the  load  is  to 
separate  the  particles  in  this  section  above  the  center  line, 
and  to  push  those  below  the  center  line  closer  together;  in 
other  words,  through  the  bending  action  of  the  load,  the 
upper  part  of  the  beam  is  subjected  to  a  tensile  stress,  while 
the  lower  is  subjected  to  a  compressive  stress. 

Fig.  54  also  shows  that  the  greater  the  distance  of  the 
particles  in  the  assumed  section  above  or  below  the  center 
x,  the  greater  will  be  their  displacement;  since  the  stress 
in  a  loaded  body  is  directly  proportional  to  the  strain,  or 
relative  displacement  of  the  particles,  it  follows  that  the 
stress  in  a  particle  of  any  section  is  proportional  to  its 
distance  from  the  center  line,  and  that  the  greatest  stress  is 
in  the  particles  composing  the  upper  and  lower  surfaces  of 
the  beam. 

In  accordance  with  the  conditions  of  equilibrium,  the 
algebraic  sum  of  the  moments  of  all  the  forces  tending  to 
produce  rotation  around  a  given  center  must  be  zero ;  we 
have  seen  that  the  weight  of  the  load  is  a  force  which  tends 
to  produce  right-hand  rotation  around  the  center  x ;  there- 
fore, if  the  beam  does  not  break  under  the  action  of  the 


76 


ARCHITECTURAL  ENGINEERING. 


load,  there  must  be  forces  acting  whose  moments,  with 
respect  to  the  center  ,r,  balance  the  moment  of  the  load. 
These  forces  are  the  resistances  with  \vhich  the  particles  of 
the  beam  oppose  any  effort  to  change  their  relative  posi- 
tions. The  tensile  stresses  in  the  particles  above  the  center 
.r,  and  the  compressive  stresses  in  those  below  it,  are  a  set 
of  forces  which  resist  the  tendency  of  the  load  to  turn  the 
end  of  the  beam,  and,  when  the  effect  of  the  load  is  just 
balanced  by  the  effect  of  these  forces,  it  is  evident  that  the 
sum  of  the  moments  of  these  resisting  stresses  is  equal  to 
the  moment  of  the  load.  The  sum  of  the  moments  of  the 
stresses  of  all  the  particles  composing  any  section  of  a  beam  is 
called  the  resisting  moment  or  moment  of  resistance  of 
that  section. 

9O.  Neutral  Axis.— In  Fig.  55,  let  A  B  CD  represent 
a  cantilever.  A  force  F  acts  upon  it;  at  its  extremity  A ; 
the  principles  developed  in  Art.  89  show  that  the  force  will 
tend  to  bend  the  beam  into  the  shape  shown  by  A'  B  CD'. 


FIG.  55. 

It  is  evident  from  what  has  preceded  that  the  upper  part 
A '  B  is  now  longer  than  it  was  before  the  force  was  applied ; 
i.  e.,  A'  R  is  longer  than  A  B.  It  is  also  evident  that  D'  C 
is  shorter  than  D  C.  Hence,  the  effect  of  the  force  F  in 
bending  the  beam  is  to  lengthen  the  upper  fibers  and  to 
shorten  the  lower  ones.  Further  consideration  will  show 
that  there  must  be  a  fiber,  S  S",  which  is  neither  lengthened 
nor  shortened  when  the  beam  is  bent,  i.  e. ,  5  S"  =  S'  S". 


ARCHITECTURAL  ENGINEERING. 


77 


When  the  beam  is  straight,  the  fiber  S  S",  which  is  neither 
lengthened  nor  shortened  when  the  beam  is  bent,  is  called 
the  neutral  line. 

The  neutral  line  corresponds  to  the  center  line  a  b,  Fig.  54, 
on  which  the  center  of  moments  x  was  taken. 

91.  The  relations  between  the  effect  of  a  load  and  the 
resulting  stresses  in  a  beam  have  been  thoroughly  proved, 
both  by  mathematical  investigations  and  numerous  experi- 
ments. The  results  of  these  experiments  on  beams  may  be 
briefly  expressed  by  the  following 

Experimental  HjBwr.—When  a  beam  is  bent,  the  horizontal 
elongation  (or  compression]  of  any  fiber  is  directly  proportional 
to  its  distance  from  the  neutral  surface,  and,  since  the  strains 
are  directly  proportional  to  the  horizontal  stresses  in  each 
fiber,  they  are  also  directly  proportional  to  their  distances 
from  the  neutral  surface,  provided  the  elastic  limit  is  not 
exceeded. 

The  line  sl  s^  which  passes  through  any  section,  as  abed, 
Fig.  55,  perpendicular  to  the  neutral  line,  is  called  the 
neutral  axis,  and  the  surface  5,  S' S"  S^  Fig.  5G,  which 


FIG.  56. 


separates  all  the  parts  of  the  beam  above  any  neutral  line, 
or  neutral  axis,  from  those  below,  is  called  the  neutral 
surface.  The  neutral  axis,  then,  is  the  line  of  intersection 


78  ARCHITECTURAL  ENGINEERING.  §  5 

of  a  cross-section  with  the  neutral  surface.  It  is  shown  in 
works  on  mechanics  that  the  neutral  axis  always  passes 
through  the  center  of  gravity  of  the  cross-section  of  the 
beam. 

92.  Bending  Moments  In  Simple  Beams. — Referring 
to  the  simple  beam  shown  in  Fig.  52,  let  us  take  the  center 
of  moments  on  the  neutral  axis  directly  under  the  load  a, 
and  consider  the  effect  produced  on  a  vertical  section  of  the 
beam  through  this  center,  by  the  reaction  Rt.     It  was  shown 
in  Art.  87  that  the  reaction  R^  is  an  upward  force  of  6,933^ 
pounds ;  it  therefore  has  a  tendency  to  turn  the  end  of  the 
beam  upwards  around  the  assumed  center  with  a  moment  of 
6,933|X  3  =  20,800  foot-pounds.      It  is  evident  that,  to  pre- 
vent this  turning  from  actually  taking  place,  the  positive 
moment  of  the  reaction  must  be  balanced  by  a  negative 
moment,  which  can  be  produced  only  by  a  set  of  internal 
stresses.     The  condition  that  the  moment  of  the  stresses 
must  be  negative  makes  it  plain  that  the  upper  fibers  must 
be  in  compression  and  the  lower  in  tension,  a  result  exactly 
opposite  to  the  effect  produced  by  the  bending  moment  on 
the  fibers  in  the  cantilever. 

93.  Effect  of  the  Moments  Due  to  Loads. — The  only 
force  acting  on  the  beam  at  the  left  of  the  section  considered 
in  Art.  92  was  the  reaction  R^     The  load  a  acted  down- 
wards directly  through  this  section,  but  its  lever  arm,  and 
consequently  its  moment,  with  respect  to  the  assumed  center, 
was  zero.     Take  now  a  point  on  the  center  line  of  the  beam 
directly  under   the   positive   load  b.      The   reaction   has  a 
moment  with  respect  to  this  center  of  6,933|  X'll  =  76,266f 
foot-pounds,  while  the  load  a,  which  acts  downwards  with  a 
lever  arm  of  8  feet,  has  a  negative  moment  of  2,000x8 
=  16,000  foot-pounds.    The  bending  moment  at  the  assumed 
section  is   the   algebraic   sum    of   these   moments,   that  is, 
76,266f  - 16,000  =  60,266f  foot-pounds.     Again,  taking  the 
center  of  moments  on  a  section  9  feet  from  the  right  reac- 
tion R  ,  the  moments  are  as  follows: 


§  5  ARCHITECTURAL   ENGINEERING.  70 

Positive  moment: 

Reaction  A',,  G,933:Vx21  =  1  4  5,  G  0  0  ft.-lb. 

Negative  moments: 

Load*,  2,000X18  =  30,000  ft.-lb. 

Load  b,  0,000X10  =  00,000  ft.-lb.        9  0,  0  0  0  ft.-lb. 

'Difference 4  9,  0  0  0  ft.-lb. 

This  resultant  moment  is  the  bending  moment  at  the  given 
section. 

94.  The  illustrations  show   that    the  bending  moment 
varies  from  point  to  point  in  a  beam,  and  depends  on  the 
length  of  the  beam  and  on   the  size  as  well  as  position  of 
the  loads.      Since  the  stresses  in  the  beam,  and  consequently 
its  ability  to  carry  its  loads,  depend  directly  on  the  bending 
moment,  it  follows  that  it  is  important  to  find,  not  only  the 
bending  moment  for  any  assumed  section,  but  also  the  sec- 
tion where  the  bending  moment  is  greatest.      It  is,  in  this 
connection,  useful  to  note  the  relation  between  the  bending 
moment  and  the  shear. 

95.  The  shear  in  a  simple  beam  is  always  greatest  at  the 
greater  reaction,  being  equal  to  that  reaction.      In  passing 
along  the  beam  from  either  reaction,  there  is  no  change  in 
the  shear  until  a  load  is  reached.    At  each  point  where  a  load 
is  added,  the  shear  is  diminished  by  an  amount  equal  to  the 
load.     At  the  point  where  the  sum  of  the  added  loads  equals 
or  exceeds  the  reaction,  the  shear  is   said  to  change  sign. 
The  section  where   the  change  in   sign   in  the  shear  takes 
place,  depends  on  the  method  of  loading.    With  a  uniformly 
distributed  load,  the   shear  diminishes  uniformly  from  each 
reaction,  and  the  section  where  the  sign  changes  is  the  sec- 
tion of  the  beam  midway  between  the   supports.     With  a 
single  concentrated  load,  the  shear  is  equal  to  each  reaction 
at  all  sections  between  that  reaction  and  the  point  where  the 
load  is  applied,  and  the   section   where  the  shear  changes 
sign  is  directly  under  the  load.    With  any  system  of  loading, 
the  section  where  the  shear  changes  sign  can  be  found  by 
adding  together  the   successive  loads  from  either  reaction 
towards  the  center  of  the  beam,   until  a  sum  is  obtained 


80 


ARCHITECTURAL  ENGINEERING. 


which  equals  or  exceeds  the  reaction ;  the  section  where  the 
shear  changes  sign  is  under  the  point  of  application  of  the 
last  load  added. 

96.  The  bending  moment  in  a  simple  beam  increases 
as  the  shear  decreases;  it  is  zero  at  either  reaction  and  in- 
creases towards  the  center,  becoming  greatest  at  the  section 
where  the  shear  changes  sign.  With  a  uniformly  distributed 
load,  the  greatest  bending  moment  is  at  the  section  of  the 
beam  midway  between  the  supports ;  with  a  single  concen- 
trated load,  the  greatest  bending  moment  is  directly  under 
the  load ;  and  with  any  system  of  loading,  the  greatest  bend- 
ing moment  occurs  at  the  section  where  the  shear  changes 
sign.  Having  located  the  section  of  the  greatest  bending 
moment,  the  magnitude  of  this  moment  can  be  readily  com- 
puted by  taking  the  center  of  moment  on  the  section  of 
greatest  bending  moment  and  computing  the  resultant 
moment  of  either  reaction  and  all  the  loads  between  it 
and  the  center  in  question. 

EXAMPLE. — A  wooden  beam,  supported  on  two  brick  piers,  is  loaded 
as  shown  in  Fig.  57.  (a)  What  is  the  greatest  shear  ?  (b)  Where  does 


FIG.  57. 


the  shear  change  sign  ?    (c)  What  is  the  greatest  bending  moment  in 
inch-pounds  ? 

SOLUTION. — (a)  Since  the  greatest  shear  is  equal  to  the  greater  reac- 
tion, we  will  first  compute  the  reactions.    Taking  the  center  of  moments 


§5  ARCHITECTURAL  ENGINEERING.  81 

at  the  edge  of  pier  a,  and  remembering  that  the  moment  of  the  uni- 
form load  is  the  same  as  the  moment  of  an  equal  concentrated  load 
acting  at  the  center  of  gravity  of  the  uniform  load,  the  moments  of  the 
loads  are: 

10,000-pound  load 10,000  X    6     =      6  0,0  0  0  ft.-lb. 

Load  d 6,000  X  IS     =      7  8,0  0  0  ft.-lb. 

Load  c 4,000X16     :         6  4,0  0  0  ft.-lb. 

Uniformly  distributed  load. .  .   12,500  X  1-i  =  1  5  6,2  5  0  ft.-lb. 
Total 3  5  8,2  5  0  ft.-lb. 

This  also  equals  the  moment  of  the  reaction  of  the  pier  /;.  The  reac- 
tion at  b  is,  therefore,  358,250  -4-  25  =  14,330  pounds.  The  total  load  is 
10,000  +  6,000  +  4,000  +  12,500  —  32,500  pounds;  the  reaction  at  a  is, 
therefore,  32,500  —  14,330  =  18,170  pounds.  This,  being  the  greater 
reaction,  is  the  greatest  shear.  Ans. 

(d)  Beginning  at  the  left  reaction  and  adding  the  loads  in  succession 
towards  the  right,  we  find  that  the  load  of  10,000  pounds  plus  the  uni- 
formly distributed  load  between  the  reaction  and  the  point  of  applica- 
tion of  the  load  d,  is  10,000  +  500  X  13  =  16,500  pounds.  This  is  less 
than  the  left  reaction,  but  when  we  add  the  load  d,  the  sum  of  the 
loads  is  greater  than  the  reaction ;  consequently,  the  shear  changes 
sign  under  the  load  d.  Ans. 

(c)  Taking  the  center  of  moments,  under  the  load  tf,  and  considering 
the  forces  at  the  left,  we  have 

Positive  moment: 

18,170  X  13  =  2  3  6,2  1  0  ft.-lb. 

Negative  moments: 

6,500  X6|  =  42,2  50  ft.-lb. 

10,000X7"  =  7  0,0  0  0  ft.-lb. 

1  1  2,2  5  0  ft.-lb. 
Difference  (bending  moment  in  ft.-lb.)     1  2  3,9  6  0  ft.-lb. 

The  greatest  bending  moment  in  inch-pounds  is,  therefore,  123,960 
X  12  =  1,487,520  inch-pounds.  Ans. 

97.     General    Formulas    for    Bending    Moments. — 

Where  a  cantilever  or  a  simple  beam  is  symmetrically 
loaded,  it  is  not  necessary  to  calculate  the  moments  of  the 
forces  acting  upon  the  beam  in  order  to  find  the  reactions 
and  bending-  moments.  The  rules  and  formulas  of  Table  10 
are  available  in  computing  the  bending  moment  J/  in  the 
five  simple  cases  of  beam  loading  set  forth. 


ARCHITECTURAL  ENGINEERING. 


CO 


Loading. 


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o   c  .5  T3 


a^ 


§5  ARCHITECTURAL  ENGINEERING.  83 

98.  The  rule  given  in  Case  IV  is  that  most  used,  as  it 
applies  to  a  beam  uniformly  loaded,  such  as  floor  joists, 
girders,  and,  in  some  cases,  the  rafters  of  a  roof.  The  rule 
in  Case  V  is  convenient  in  calculating  the  bending  moment 
on  lintels  supporting  brickwork  or  masonry  over  openings. 
If,  in  Case  III,  the  beam  is  firmly  fixed  or  fastened  at  both 
ends,  the  bending  moment  under  the  same  load  will  be  only 
half  as  much.  If,  in  Case  IV,  the  ends  of  the  beam  are 
firmly  fixed,  instead  of  dividing  by  8  a  constant  of  12  is  to 
be  used.  It  is  seldom  advisable,  in  ordinary  building  prac- 
tice, to  consider  the  ends  of  a  beam  fixed,  it  being  good 
practice  to  assume  the  ends  of  the  beams  as  simply  bearing 
on  the  wall,  using  the  rules  and  formulas  in  Table  10.  It 
should  be,  however,  understood  that  all  these  rules  and 
formulas  apply  to  static  loads.  The  same  load  suddenly 
applied  produces  a  stress  in  the  beam  twice  as  great  as  that 
of  a  static  load.  The  safe  sudden  load  is,  therefore,  only 
half  as  much. 

ExAMi'i.K. — What  will  be  the  bending  moment  in  inch-pounds  on  a 
wood  girder  supporting  a  floor  area  of  150  square  feet,  the  dead  and 
live  load  being  100  pounds  per  square  foot,  and  the  span  of  the  girder 
20  feet  ? 

SOLUTION. — The  total  uniformly  distributed  load  is  150  X  100 
=  15,000  pounds ;  therefore,  by  applying  the  formula  in  Case  IV, 

_,       II' L        15,000X20X12 
lablelO,  we  have  the  bending  moment  J/  =  — ^ —  =  „ — 

o  o 

=  450,000  inch-pounds.     Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  A  beam  has  a  span  of  20  feet,  and  is  loaded  with  a  uniformly 
distributed  load  of  2,500  pounds  per  lineal  foot.     What  is  the  greatest 
bending  moment  in  inch-pounds  upon  the  beam  ?    Ans.  1,500,000  in. -Ib. 

2.  What  is  the  bending  moment  in  foot-pounds  on  a  cantilever 
beam  securely  fastened  into  a  wall,  extending  from  the  point  of  support 
10  feet,  and  loaded  with  a  uniformly  distributed  load  of  1,000  pounds 
per  lineal  foot  ?  Ans.  5,000  ft.-lb. 

3.  What  is  the  bending  moment  in  inch-pounds  on  a  girder  having 
a  span  of  30  feet,  if  there  is  a  uniformly  distributed  load  of  1,500 
pounds  per  lineal  foot,  and  a  load  of  20,000  pounds  concentrated  at 
the  center?  Ans.  3,825,000  in.-lb. 

1-18 


84  ARCHITECTURAL  ENGINEERING.  §  5 

4.  A  plate  girder  in  a  building  is  required  to  support  a  uniformly 
distributed  load  of  2,000  pounds  per  lineal  foot,  extending  20  feet  each 
side  of  the  center  of  the  girder;  in  addition,  it  is  required  to  support  a 
load  of  80,600  pounds,  concentrated  10  feet  from  one  end  of  the  girder, 
and  another  load  of  43,000  pounds,  located  22  feet  from  the  same  end. 
What  will  be  the  greatest  bending  moment  en  the  girder,  in  foot- 
pounds, if  the  span  is  60  feet  ?  Ans.  1,535,000  ft.-lb. 


STRENGTH    OF    BEAMS. 

99.  Resisting  Moment. — It  was  stated  in  Art.  89,  that 
the  resisting"  moment  of  any  section  of  a  beam  is  the  sum  of 
the  moments  of  all  the  stresses  produced  in  that  section  by 
the  bending  moment ;  it  was  also  shown  that  the  resisting 
moment  must  equal  the  bending  moment.      By  higher  math- 
ematics it  is  proved  that  the  resisting  moment  is  equal  to 
the  product  of  the  greatest  unit  stress  in  any  part  of  a  sec- 
tion multiplied  by  a  factor,  called  the  section  modulus,  or 
the  resisting  inches,  which  depends  on  the  shape  of  the 
section. 

100.  If  we  ..assume  the  greatest  unit  stress  to  be  the 
modulus  of  rupture  of  the  material  composing  a  beam,  we 
have  the  following  rule : 

Rule. —  To  find  the  ultimate  resisting  moment  of  a  beam, 
multiply  the  section  modulus  by  the  modulus  of  rupture  of  the 
material  of  which  the  beam  is  composed. 

The  modulus  of  rupture  for  the  materials  used  in  building 
construction  may  be  obtained  from  Table  6. 

101.  Section  Modulus. — The  general  method  for  find- 
ing the  section  modulus  of  any  section  of  a  beam  will  not  be 
discussed  here,  but  a  number  of  rules,  formulas,  and  tables 
are  given  from  which  to  find  the  section  modulus  of  the  sec- 
tions used  in  ordinary  building  operations. 

Rule. —  To  obtain  the  section  modtilus  of  any  rectangular 
beam,  multiply  the  square  of  its  depth  in  inches  by  its  width 
in  inches  and  divide  by  6. 


§5  ARCHITECTURAL  ENGINEERING.  85 

This  rule  is  expressed  by  the  formula 
/'       bd*  „ 

K    :       —  ,  (li>.) 

in  which  K  =  section  modulus  ; 

d  =  depth  of  beam  in  inches; 

/;    =  breadth  or  width  of  beam  in  inches. 

EXAMPLE  1.  —  (a)  What  will  be  the  section  modulus  of  a  yellow-pine 
beam,  10  inches  wide  by  12  inches  in  depth  ?  (b)  What  will  be  the  ulti- 
mate resisting  moment  of  this  beam  ? 

SOLUTION.  —  (a)  Here          d—  12  inches; 
b  =  10  inches. 

Therefore,  by  applying  the  formula,  we  have 

A'=*£  .=  1^==  1^  =  240.     An, 

(&)  According  to  Table  6,  the  modulus  of  rupture  for  yellow  pine  is  7,300 
pounds  per  square  inch  ;  hence,  the  resisting  moment  of  the  10"  X  12" 
yellow-pine  beam  is  240  X  7,800  =  1,752,000  inch-pounds.  Ans. 

EXAMPLE  2.  —  What  size  of  spruce  girder  is  required  to  support  a 
uniformly  distributed  load  of  500  pounds  per  lineal  foot,  the  span  of 
the  girder  being  22  feet,  and  the  factor  of  safety  4  ? 

SOLUTION.  —  First  find  the  bending  moment  in  inch-pounds.  The 
total  load  on  the  girder  is  500  X  22  =  11,000  pounds.  The  bending 
moment  due  to  a  uniformly  distributed  load,  according  to  Table  10,  is 

W  T  11  000  V  22 

—  g-.      Then  --  ^  -  =  30,250   foot-pounds,   which   multiplied   by 

12  gives  363,000,  the  bending  moment  in  inch-pounds.  To  find  the  size 
of  a  spruce  beam  having  a  safe  resisting  moment  equal  to  this  bending 
moment,  take  a  10"  X  14"  beam  —  the  section  modulus  from  formula  15 
being 

bd*       10  X  142 

K  ~-  T"       T" 

The  modulus  of  rupture  of  spruce  being  4,800,  the  ultimate  resisting 
moment  of  the  beam  is  326  X  4,800  =  1,564,800  inch-pounds.  If  a 
factor  of  safety  of  4  is  used,  the  safe  resisting  moment  of  the  beam  is 
1,564,800-7-4  =  391,200  inch-pounds.  Since  the  bending  moment  is 
only  363,000  inch-pounds,  the  beam  is  amply  strong.  Ans. 

1O2.  To  determine  the  uniformly  distributed  load  that 
will  break  a  beam  whose  size  is  known  take  the  formula 


12 
—      =  -W7-,        (16.) 


8G  ARCHITECTURAL  ENGINEERING.  §  5 

where        K  —  the  section  modulus  of  the  beam  ; 

.V  =  modulus  of  rupture  of  the  material; 

L  =  span  of  beam  in  feet  ; 

B  =  breaking  load  of  rectangular  beam. 

This  formula  may  be  stated  by  the  following  rule  : 

.Rule.  —  To  determine  t/ie  breaking  load  in  pounds  of  any 
rectangular  beam,  multiply  the  sectional  modulus  of  tJie  beam 
by  the  modulus  of  rupture  of  tJie  material  ;  multiply  this 
product  by  2  and  divide  the  result  by  3  times  the  span  of  the 
beam  in  feet. 

EXAMPLE.  —  (a)  What  uniformly  distributed  load  will  break  a  hemlock 
joist,  3  inches  by  14  inches,  the  span  being  25  feet  ?  (I))  What  will  be 
the  safe  load  if  a  factor  of  safety  of  4  is  used  ? 

b  d*        3  X  142 
SOLUTION.  —  (a)  Section  modulus  =  -^—  =  -  —  ^  —  =  98.  The  modu- 

lus of  rupture  for  hemlock,  according  to  Table  6,  is  3,600  pounds  per 
square  inch.     Substituting  these  values  in  the  formula,  we  have 

2  X  98  X  3,600 


3x^5 


=  9,408  pounds.     Ans. 


(fi)  If  a  factor  of  safety  of  4  is  used,  the  safe  uniformly  distributed 
load  that  this  beam  will  carry  is  9,408  pounds  -f-  4  =  2,352  pounds.  Ans. 

1O3.  Steel  ]Jeams.  —  Steel  beams  used  in  building  con- 
struction may  be  either  channels,  I  beams,  angles,  or 

tees.  For  floorbeams,  channels  and  I  beams  are  principally 
used,  while  angles  and  tees  are  used  in  roof  trusses  and  like 
construction.  The  strength  of  channels  and  I  beams  only 
are  here  to  be  considered. 

In  engineering,  the  two  rolled  shapes,  channels  and  I 
beams,  are  generally  called  channels  and  beams.  When  the 
word  beam  is  used,  it  is  generally  understood  that  an  I 
beam  is  meant.  For  instance,  a  12-inch,  40-pound  beam 
would  mean  an  I  beam  12  inches  in  depth,  weighing  40 
pounds  to  the  lineal  foot  ;  while  a  channel  of  the  same  size 
would  be  expressed  as  a  12-inch,  40-pound  channel.  In  des- 
ignating rolled  shapes  on  working  drawings,  various  sys- 
tems of  abbreviations  are  used.  A  12-inch,  40-pound  beam 
may  be  expressed  as  12"  I  40$,  or  a  channel  as  12"  C  40#- 


§  5  ARCHITECTURAL  ENGINEERING.  87 

This  is  entirely  a  matter  of  judgment  with  the  draftsman, 
or  is  governed  by  the  practiee  used  in  the  particular  draft- 
ing room.  As  long  as  the  size,  character,  and  weight  of 
the  beam  are  given,  it  matters  little  how  expressed,  if  intel- 
ligibly written. 

104.  Strength  of  Steel  Beams. — In    calculating    the 
strength  of  steel   beams,    it  is  first    necessary  to    find    the 
breaking  moment,  using  the  methods  and  rules  already  laid 
down.    Then  the  section  modulus  required  in  the  beam  may 
be  obtained  by  dividing  the  bending  moment  in  inch-pounds 
by  the  quotient  obtained  by  dividing  the  modulus  of  rup- 
ture by  the   factor  of  safety.      Assume,    for  example,    the 
bending   moment    on    a   beam    to    be    50,000    foot-pounds. 
Reduce   it  to  inch-pounds  by  multiplying  it  by  12,  which 
gives  600,000  inch-pounds.      The  modulus  of  rupture   for 
structural  steel  is  60,000  pounds.      If  a  factor  of  safety  of  4 
is  used,    the  safe  working   value   of  this  material    will   be 
60,000-^-4  =  15,000  per  square  inch.    Then  000,000-4-15,000 
=  40,  the  section  modulus  required. 

105.  The  approximate  section  modulus  of  an  I  beam 
or  a  channel  may  be  found  by  the  following  rules : 

Rule  I. —  To  obtain  the  approximate  section  modulus  of  an 
I  beam,  multiply  the  sectional  area  of  the  beam  in  square 
inches  by  the  depth  in  inches,  and  divide  by  the  constant  3.2. 

Rule  II. —  To  obtain  the  approximate  section  modulus  of  a 
channel,  multiply  the  sectional  area  of  the  channel  in  square 
inches  by  the  depth  in  inches,  and  divide  by  the  constant  3.67. 

Letting  a  =  the  sectional  area  of  an  I  beam  or  a  channel 
in  square  inches,  and  h  its  depth  in  inches,  the  approximate 
section  modulus  of  an  I  beam  may  be  found  from  the 
formula 

K>  =  o- 

and  of  a  channel,     Kc  =  — — .         (18.) 

o.  b7 

EXAMPLE. — What  is  the  section  modulus  of  a  12-inch  I  beam,  the 
sectional  area  of  which  is  9.01  square  inches  ? 


88  ARCHITECTURAL  ENGINEERING.  §  5 

SOLUTION. — Applying  the  formula,  we  have 
K_ah_  9.0002  _ 
Ki  -  O  -    ~3^~ 

106.  Tables  11  and  12  give  the  principal  elements  of  I 
beams  and  channels,  and  obviate  the  necessity  of  calculating 
the  section  modulus,  this  being  given    under   the    column 
headed  "  Section  Modulus  on  Axis  A  B" 

The  moment  of  inertia  on  the  axis  A  B  is  also  given  in 
these  tables.  The  section  modulus  of  any  beam  may  be 
obtained  by  dividing  the  moment  of  inertia  of  the  beam  sec- 
tion by  one-half  the  depth  of  the  beam  in  inches.  In  Table 
11,  for  instance,  "Elements  of  I  Beams,"  a  15-inch,  69.2- 
pound  beam  has  a  moment  of  inertia  of  710.  If  this  is 
divided  by  one-half  the  depth  of  the  beam  in  inches,  in  this 
case  7-|,  the  result  obtained  will  be  94. 66f,  which,  as  may  be 
seen  by  referring  to  the  column  headed  "Section  Modulus 
on  A  B"  is  the  correct  section  modulus  of  this  beam. 

107.  To  illustrate  the  method  of  calculating  the  dimen- 
sions of  a  steel  beam,  let  it  be  required  to  find  what  size  of 
steel  I  beams  is  necessary  to  support  the  floor  of  an  office 
building,  this  floor  resting  on  brick  arches  sprung  between 
the  beams  and  weighing  complete  110  pounds  per  square 
foot.     The  building  is  designed  to  carry  a  live  load  of  40 
pounds  per  square  foot.     The  span  of  the  beams  is  20  feet, 
spaced  5  feet  on  centers.     The   owner   requires    that   the 
building  have  a  large  factor  of  safety,  and  suggests  that  for 
the  floorbeams  a  safety  factor  of  5  be  used.     The  total  dead 
and  live  load  on  the  floor  is  110  Ib.  +40  Ib.   —  150  pounds 
per  square  foot.     The  floor  area  supported  by  one  beam  is 
20  X  5  —  100  square  feet.     Then  the  total  load  on  one  beam 
is  100x150  =  15,000  pounds.     The  load   being  uniformly 
distributed,  the  formula  for  the  bending  moment,  according 

W I 
to  Table  10,  is  M  =  -    — ;    substituting  the   values  for    W 

o 

and  L,  M  -     °'        *    -  =  37,500,  the  bending  moment  in 

foot-pounds,  which,  being  multiplied  by  12,   gives  450,000 
inch-pounds. 


TABLE   11. 
ELEMENTS  OF  I  BEAMS. 


Size  in 
Inches. 

Area  in 
Square 
Inches. 

Weight  in 
Pounds  pel- 
Foot. 

Moments  of 
Inertia  on 
Axis  A  B. 

Section 
Modulus  on 
Axis  A  B. 

3 

1.56 

5-3 

2.41 

I.6l 

3 

2.OI 

6.8 

2-75 

1-83 

4 

I.  Si 

6.2 

5.02 

2-51 

4 

2.17 

7-4 

5-82 

2.91 

5 

2.76 

9-4 

11.58 

4-63 

5 

3.6l 

12.3 

13-34 

5-34 

6 

3-51 

11.9 

21.14 

7-05 

6 

4-47 

15-2 

24.02 

8.01 

6 

9-49 

32.3 

52.53 

17-51 

6 

10.99 

37-4 

57-03 

19.01 

6 

12.06 

41.0 

64.07 

21.36 

6 

I3-56 

46.  i 

68.57 

22.86 

7 

4-31 

14.6 

35-43 

IO.  12 

.     7 

5-29 

17.9 

39-43 

11.27 

8 

5-12 

17.4 

54-31 

13.58 

8 

6.24 

21.2 

60.28 

15-07 

9 

6.04 

20.5 

80.78 

17-95 

9 

7.48 

25-4 

90.50 

2O.  II 

10 

6.91 

23-5 

112.42 

22.48 

10 

8.78 

29.8 

141.44 

28.29 

10 

8.91 

30-3 

129.08 

25.82 

IO 

10.28 

34-9 

153-94 

30-79 

12 

9.01 

30.6 

207.90 

34-65 

12 

11.29 

38.4 

233.80 

38.97 

12 

n.  60 

39-4 

268.30 

44-72 

12 

14.00 

47.6 

299.76 

49.96 

12 

16.32 

55-5 

362.88 

60.48 

12 

19.68 

66.9 

403-38 

67.23 

15 

I2.II 

41.2 

433-00 

57-73 

15 

14.49 

49-3 

518.61 

69.15 

15 

15.56 

52.9 

497.68 

66.36 

15 

16.74 

56.9 

560.  79 

74-77 

15 

16.95 

57-6 

583-78 

77-84 

15 

20.38 

69.2 

710.00 

94-67 

15 

2O.7O 

70.4 

654.09 

87.21 

15     . 

25.03 

85.1 

789-24 

105.23 

20 

19.04 

64.8 

1,145-79 

114-58 

20 

22.94 

78.0 

1,367.37 

136.74 

20 

24.04 

81.7 

1,312.46 

131.24 

20 

28.94 

98.4 

i,567.37 

156.74 

TABLE   12. 
ELEMENTS  OF  CHANNELS. 


A 


J 


Area  in 
Size  in            Square 
Inches.            Inches. 

Weight  in 
Pounds  per 
Foot. 

Moments  of 
Inertia  on 
Axis  A  B. 

Section 
Modulus  on 
Axis  AB. 

If 

0-33                           LI 

0.14 

0.  16 

2 

0.87 

2.97 

0.48 

0.48 

2 

1.  06 

3.60 

0.52 

0.52 

21 

1.  12 

3-8 

o.So 

0-35 

3 

I-5I 

5-i 

2.04 

1.36 

3 

I.73 

6.1 

2.24 

1.49 

4 

1.58 

5-4 

3-6? 

1.84 

4 

2.30 

7-8 

4.64 

2.32 

5 

i-93 

6.6 

6.64 

2.66 

5 

2.83 

9.6 

8.52 

3-41 

6 

2.27 

7-7 

11.62 

3.87 

6 

3-24 

II.  O 

17-54 

5-84 

6 

3-35 

11.4 

14.86 

4-95 

6 

5.46 

18.6 

24.20 

8.06 

7 

2.67 

9.1 

19-39 

5-54 

7 

3-75 

12.8 

27.14 

7-75 

7 

3.86 

13-1 

24-25 

6-93 

7 

7.'n 

24.2 

40.86 

11.67 

8 

3.22 

10.9 

29.76 

7-44 

8 

4.29 

14.6 

39.76 

9.94 

8 

4.42 

15-0 

36.16 

9.04 

8 

6.05 

20.6 

49-15 

12.28 

9 

3-94 

13.4 

46.48 

10-33 

9 

5-17 

17.6 

59.89 

13-31 

9 

5-74 

19.5 

58.63 

13-03 

9 

7.96 

27.1 

78.72 

17-49 

10 

4.84 

16.5 

71.09 

14.22 

10 

6.17 

20.9 

88.01 

17.60 

10 

7.24 

24.6 

91.09 

18.62 

10 

10.37 

35.3 

123.01 

24.60 

12 

5-8i 

19.8 

118.32 

19.72 

12 

6.07 

20.  6 

123.38 

20.56 

12 

9.17 

31.2 

157-20 

26.20 

12 

9.24 

31-4 

189.93 

31-65 

12 

9-43 

32.1 

163.71 

27.28 

12 

16.32 

55-5 

274.89 

45.81 

15 
15 
15 

9.62 
14.64 

14.87 

32-7 
49.8 
50.6 

286.84 
435-72 
385-28 

38.25 
58.09 
51-30 

15 

20.34 

69.2 

542.59 

72.34 

§  5  ARCHITECTURAL  ENGINEERING  91 

The  modulus  of  rupture  for  structural  steel  being  60,000 
pounds  per  square  inch,  and,  since  a  factor  of  safety  of  5  is 
required,  the  safe  working- value  will  be  60. 000-=- 5  =  12,000 
pounds  per  square  inch.  The  bending  moment  in  inch- 
pounds  is  450,000,  which  divided  by  12,000  gives  a  section 
modulus  of  37.5. 

Referring  to  Table  11,  it  is  seen  that  the  section  modulus 
on  the  axis  A  />  of  a  12-inch,  39.4-poimd  beam  is  4-4.72, 
which  is  ample,  for  a  modulus  of  37.5  only  is  required. 

Referring,  also,  to  the  15-inch  beam  in  the  same  table,  it 
is  seen  that  the  section  modulus  of  a  15-inch  beam,  41.2 
pounds,  is  57. 73,  and,  as  it  weighs  not  2  pounds  more  per 
foot,  it  might  be  advisable  to  use  this  beam,  especially  as 
the  owner  requires  a  large  factor  of  strength. 

In  selecting  beams  from  the  table,  care  should  be  taken  to 
obtain  the  deepest  beams  of  the  least  weight  with  the 
required  section  modulus.  Thus,  by  referring  to  Table  1 1 , 
it  is  seen  that  the  section  modulus  of  a  10-inch  beam,  30.3 
pounds,  is  25.82,  while  a  12-inch  beam  of  30.6  pounds,  of 
nearly  the  same  weight  as  the  10-inch  beam,  has  a  modulus 
of  34.65,  and,  in  consequence,  possesses  nearly  one-third 
more  strength,  making  it,  therefore,  the  more  economical 
beam  to  use. 

1O8.  By  way  of  general  review  of  the  subject  of  beams, 
we  present  the  following  practical  example :  Fig.  58  shows 
the  transverse  sectional  elevation  of  a  large  department 
store.  It  will  require  two  I  beams  to  form  the  girder  B. 
What  wrill  be  the  size  and  weight  of  these  steel  beams  ? 

Before  commencing  the  calculations,  draw  the  outline  dia- 
gram as  shown  in  Fig.  59.  These  are  called  frame  diagrams. 
The  two  supports  for  the  girder  are  the  wall  JFand  the 
column  C.  The  loads  upon  the  girder  are  the  two  uniform 
loads  g  and  h.  The  load  h  is  due  to  the  weight  of  the  floor, 
girder,  and  the  ceiling,  together  with  the  live  load  on  the 
floor,  due  to  the  people,  furniture,  etc.  This  load  has  been 
assumed  to  amount  to  500  pounds  per  running  foot  of  the 
girder.  The  load  g  being  due  only  to  the  ceiling  and  a  portion 


ARCHITECTURAL  ENGINEERING. 


of  the  roof,  and  there  being  no  floor  load  upon  it,  has  been 
considered  as  amounting  to  200  pounds  per  running  foot. 

The  girder  is  also  loaded  with  four  concentrated  loads :  a 
of  10,000  pounds,  due  to  the  weight  of  the  light  wall  and  a 
portion  of  the  roof;  d  of  20,000  pounds,  due  to  the  load 
coming  down  the  small  column  from  a  portion  of  the  roof ; 
and  two  hanging  loads  /and  e,  of  3,000  and  2,000  pounds, 
respectively,  from  the  weight  of  the  stair  landing  or  hall. 


FIG.  58. 

Let  us  now  calculate  the  reactions.     The  moments  about 

IV,  due  to  the  various  loads,  are  as  follows: 

FT.-LB. 

3,6  0  0 
2  8  0,0  0  0 
6  0,0  00 
2  7,0  00 
6  8  0,0  0  0 
6  8,0  0  0 


Load^(200x   6=     1,200  lb.). ..   1,200 X   3  = 
Load // (500  X  28  =  14,000  lb.). .  .14,000x20  = 

Load  a .10, 000  X    6  = 

Load/ 3,OOOX    9  = 

Load  d 20,000X34  = 

Load  e 2,000  X  34  = 


Total 1,1  1  8,6  00 


ARCHITECTURAL  ENGINEERING. 


This,  divided  by  the  distance  between  the  supports,  or  the 
span,  25  feet,  gives  4-4,744:,  the  load  in  pounds  coming  on 


Ib.per  running  ft.fiOO  lb.  per  running  ft.  Beam  B 


—9ft- 


f=300O  lb. 


-34  ftr 


-25ft- 


-9ft- 


Fio.  59. 


the  column  C;  or,  in  other  words,  the  reaction  at  C. 
loads  are  as  follows  : 


The 


-=  l,2001b. 
Load  h  —  1  4,  0  0  0  lb. 
Load  a  —  1  0,0  0  0  lb. 
Load/=  3,0001b. 
Load  d  =  2  0,0  0  0  lb. 
Load  e  —  2,0  0  0  lb. 

Total  load  =  5  0,2  0  0  lb. 

Then,  the  reaction  at  W\&  50,200  —  44,744  =  5,  456  pounds. 

We  next  find  the  point  between  the  two  supports  JFand 
C,  where  the  shear  changes  sign.  Starting  from  llr,  the  first 
load  encountered  and  to  be  deducted  from  the  reaction  JFis 
the  uniform  load  g,  equal  to  200  X  6  =  1,200  pounds.  Then, 
5,456  (reaction  at  W)  —  1,200  (load  g)  =  4,250  pounds. 
The  next  load  on  the  beam  is  the  concentrated  load  a  of 
10,000  pounds,  which  is  much  more  than  the  remaining  por- 
tion of  the  reaction  IV.  The  greatest  bending  moment 
occurring  between  the  columns  and  the  wall  is,  therefore, 
at  the  point  a,  and  is  equal  to  5,456  (reaction  at  IV]  X  6 
=  32,736  foot-pounds,  less  the  moment  of  the  load  g  of 


94  ARCHITECTURAL  ENGINEERING.  §  5 

1,200X3  =  3,600  foot-pounds,  or  32,736-3,600  =  29,136 
foot-pounds. 

Again  referring  to  the  diagram,  Fig.  59,  it  is  seen  that 
there  is  a  large  bending  moment  directly  over  the  column  C, 
due  to  the  two  concentrated  loads  d  and  e  on  the  end  of  the 
beam  and  the  portion  of  the  uniform  load  //  overhanging  the 
support  C.  This  portion  of  the  beam  may  be  considered  as 
a  cantilever;  the  bending  moment  at  C  is  equal  to  the  sum 
of  the  moments  of  all  the  loads  on  the  overhanging  portion 
of  the  beam,  which  are:  Load  d,  20,000x9  =  180,000  foot- 
pounds. Load  e,  2,000x9  =  18,000  foot-pounds.  Load  h 
(overhanging  portion  —  500x9  =  4,500  pounds)  =  4,500 
X4.5  =  20, 250 foot-pounds.  Total:  218,250  foot-pounds,  or 
218,250x12  =  2,619,000  inch-pounds.  Since  this  bending 
moment  is  greater  than  that  under  the  load  a,  it  is  used  in 
determining  the  size  of  the  beam.  This  bending  moment 
divided  by  20,000,  the  safe  working  value  of  structural  steel 
(using  the  modulus  of  rupture  of  60,000  pounds -^3,  the 
safety  factor  used  in  this  case),  gives  131,  the  required  sec- 
tion modulus  in  the  two  beams.  Then,  the  section  modulus 
required  in  one  of  tlje  beams  is  131  -j-2  =  65.5. 

Referring  to  Table  11,  it  is  seen  that  the  section  modulus 
of  a  15-inch  beam,  49.3  pounds,  is  69.15.  While  this  is  in 
excess  of  the  required  amount,  it  is,  in  this  case,  the  most 
economical  beam  to  use,  and  two  of  this  kind  are  required. 

109.  Relation  Between  Span  and  Depth  of  Beam. — 

In  order  to  select  beams  that  will  not  deflect  too  much 
under  the  load  they  are  required  to  sustain,  the  depth  of 
the  beam  in  inches  should  never  be  less  than  half  the  span 
of  the  beam  in  feet.  Thus,  if  the  span  of  the  beam  be  20 
feet,  a  beam  not  less  than  10  inches  in  depth  should  be 
used,  to  avoid  excessive  deflection. 

110.  Separators  for  I  Beams. — In  building  construc- 
tion, it  frequently  happens  that  a  single  I  beam  is  insufficient 
to  carry  the  imposed  load.    Where  heavy  loads,  such  as  brick 
walls,  vaults,  etc.,  are  to  be  supported,  a  single  I  beam  is 


ARCHITECTURAL  ENGINEERING. 


inadequate,  and  two  or  more  beams  are  applied  side  by  side, 
bolted  together  with  cast-iron  or  steel  separators,  shown  in 
Fig.  60.  These  separators  hold  the  compression  flanges  of 


FIG.  60. 


the  beams  in  position,  preventing  deflection  sideways,  and 
also,  in  a  measure,  cause  the  beams  to  act  together,  dis- 
tributing the  load  uniformly  on  both.  Separators  should  be 


FIG.  61. 


spaced  from  6  to  7  feet  throughout  the  length  of  the  beam ; 
they  should  also  be  provided  at  the  supports  and  at  points 
where  heavy  loads  are  concentrated. 


96 


ARCHITECTURAL  ENGINEERING. 


111.  Beam  Girders.  —  In  designing  floors  of  buildings 
it  is  desirable  to  have  a  minimum  number  of  interior  sup- 
porting columns,  consistent  with  economy.  A  beam  girder 
consisting  of  a  pair  of  I  beams  is  frequently  advantageous 
for  supporting  the  steel  floorbeams  as  shown  at  a  in  Fig.  61. 

Girders  composed  of  two  or  more  I  beams  are  commonly 
used  to  span  openings  in  brick  walls.  If  the  wall  to  be  sup- 
ported is  thoroughly  seasoned  and  without  openings,  the 
weight  carried  by  the  girder  can  safely  be  assumed  as  the 
weight  of  a  triangular  piece  of  brickwork,  whose  altitude  is 
one-third  of  the  span  of  the  girder.  If  the  wall  is  newly 
built,  or  has  openings  for  windows  or  other  purposes,  the 
girder  must  be  designed  to  carry  the  entire  wall  above  the 
girder  between  the  supports. 

EXAMPLE.  —  Required,  the  size  of  a  steel  I  beam  girder  to  carry  a 
wall  12  inches  thick,  made  of  hard  brick  laid  in  lime  mortar;  there  are 
no  openings  in  the  wall  above  the  girder,  nor  does  the  wall  support 
floor  joists  or  roof  beams,  while  the  span  of  the  opening  is  24  feet. 

SOLUTION.  —  Draw  the  diagram  as  shown  in  Fig.  62.  The  area  of 
the  triangular  piece  of  brickwork  is  24  X  4  —  96  square  feet.  The  area 


i°fi  Span 


of  a  triangle  is  equal  to  one-half  of  the  product  of  the  base  and  altitude. 
As  the  wall  is  1  foot  thick,  there  are  96  cubic  feet  in  this  triangular 
piece.  The  weight  of  brickwork  in  lime  mortar  per  cubic  foot,  accord- 
ing to  Table  1,  is  120  pounds.  Th'en,  the  load  on  the  girder  is 
96  X  120  =  11,520  pounds. 


§  5  ARCHITECTURAL  ENGINEERING.  97 

The  bending  moment  may  be  determined  by  the  formula  or  rule 
given  in  Table  10,  for  a  beam  carrying  a  triangular  load,  or  it  may  be 
determined  by  calculating  the  moments,  as  follows:  The  reactions  at 
the  two  supports  are  each  equal  to  half  the  load,  or  11,  520  -H  2  =  5,760 
pounds.  The  greatest  bending  moment  is  at  the  center  of  the  beam. 
Then  the  moment  of  the  reaction  about  the  point  a  is  5,760  X  13 
=  69,120  foot-pounds.  But  counterbalancing  this,  and  to  be  deducted 
from  it,  is  the  moment  of  the  load  at  the  left  of  .r,  equal  to  half  of  the 
triangular  piece  of  brickwork.  The  moment  of  this  load  about  the 
point  x  is  equal  to  the  product  of  its  weight  multiplied  by  the  horizon- 
tal distance  from  a  vertical  line  through  its  center  of  gravity  to  the 
point  x.  Take  the  line  a  b  as  the  base  of  a  triangle,  remembering  that 
a  line  drawn  parallel  to  the  base  line  of  a  triangle,  at  a  distance  of 
one-third  of  the  altitude  from  it,  always  passes  through  its  center  of 
gravity.  Now  the  distance  from  the  point  x  to  the  vertical  line  through 
the  center  of  gravity  m  of  the  triangle  is  4  feet,  and  the  moment  due  to 
the  triangular  piece  of  brickwork  to  the  left  of  the  center  is  5,760x4 
=  23,040  foot-pounds.  Deducting  this  from  the  moment  of  the  reaction 
already  found,  the  moment  at  the  center  is:  69,120  —  23,040  =  46,080 
foot-pounds,  the  bending  moment  on  this  beam  or  girder;  or,  46,080 
X  12  =  552,960  inch-pounds.  This  calculation  may  be  checked  by 

W  L 
applying   the   formula  —  ^—  ^.      The   bending    moment   in   inch-pounds 

being  552,960,  using  a  safe  working  value,  or  fiber  stress,  of  15,000 
pounds,  the  section  modulus  required  is  552,960  -f-  15,000  =  36.8.  Refer- 
ring to  Table  11,  it  is  seen  that  the  section  modulus  of  a  12-inch  beam, 
38.4  pounds,  is  38.97,  which  gives  the  required  strength  in  this  case. 
It  may  be  found  to  be  better  practice  to  use  two  channels  instead  of 
one  I  beam,  for  the  top  flange  of  the  I  beam  may  be  too  narrow  to 
properly  support  the  brick  wall,  while  the  two  channels  placed  side  by 
side,  with  separators  between,  could  be  made  of  the  same  thickness  as 
the  wall. 


Connection  Angles.  —  The  standard  connection 
angles  for  the  principal  sizes  and  weights  of  steel  I  beams 
are  illustrated  in  Table  13.  These  connections  are  based 
upon  shearing  stresses  of  10,000  pounds  per  square  inch, 
bearing  stresses  of  20,000  pounds  per  square  inch,  and  an 
extreme  fiber  stress  of  10,000  pounds.  The  connections  are 
properly  designed  for  beams  whose  spans  are  not  less  than 
those  given  in  Table  14. 

When  beams  are  framed  opposite  one  another,  into  another 
beam  or  girder,  with  a  web  less  in  thickness  than  -fa  inch, 
the  minimum  lengths  of  spans  given  in  Table  14  ought  to 


98 


ARCHITECTURAL  ENGINEERING. 


TABLE    13. 

STANDARD  FRAMING   FOR  I  BEAM   CONNECTIONS. 


2  Angles  6X3£x£xo'lO4 


W'and  9" 

r2^r2t\  "'"~"' 


2  Angles 


§5  ARCHITECTURAL  ENGINEERING.  90 

be  increased  in  the  same  proportion  that  the  thickness  of  the 
web  is  to  y'V  inch.  TABLK    14. 

The      connections      in 

./       .^    .         .} 

Table    13    are    designed  ^  ,  ^-^  £ 

for     -j-f-inch     holes    and  ^  -—  ,2  •   "-^        $       "-^ 

£-inch  diameter  rivets  or  --  ^  •- 

bolts.    Connection  angles  ^  '§5.  *  'i  E.        *    •   .i  H, 

may,    if  so   specified,    be  ^  ^   '  x  ^   '        ^       ^   ' 


20      17.5      12      12.0       8        5.0 

to  the  beams;  but,  unless 

otherwise  ordered,  bolted  -°  1(;-°  ^  !'-°  r  4-° 
connections  are  generally  ir>  14.5  i^  7.5  <;  i;.o 

used.  ,  .        -         .  . 

15      12.0       10        <s.  o        a        i.o 
These  connections  are 

so  proportional  as  to  15  lu-()  10  '  7-°  4  :5-° 
cover  most  cases  occur-  15  «».o  o  i  5.5  3  ;  ^.o 
ring  in  ordinary  practice  ; 

but  where  beams  have  short  spans  and  are  loaded  to  their  full 
capacity,  connections  having  a  greater  number  of  bolts  than 
used  in  the  standard  connections  may  be  found  necessary. 

113.  Stone  Beams.  —  The  strength  of  stone  beams,  or 
lintels,  in  which  form  stone  beams  are  usually  found,  may 
be  calculated  in  the  same  manner  as  rectangular  beams  of 
any  material,  except  that  it  is  usual  to  use  a  factor  of  safety 
of  10.  First  compute  the  bending  moment  on  the  beam  in 
inch-pounds;  then  estimate  the  resisting  moment  of  the 
beam,  using  the  modulus  of  rupture  given  in  Table  0, 
which  should  be  divided  by  a  safety  factor  of  at  least  1<>. 

Another  method,  which  may  be  found  more  convenient, 
is  to  use  the  formula 

W.=  b--xc,  (19.) 


in  which      W  =  safe  uniformly  distributed  load  in  tons  of 

2,000  pounds; 

b    =  breadth  of  beam  in  inches  ; 
.  d    =  depth  of  beam  in  inches; 
/     =  span  of  beam  in  inches; 
c     —  a  coefficient  taken  from  the  following  table  : 

1-19 


100          ARCHITECTURAL  ENGINEERING.  §  5 

Coefficient  c. 

Bluestone 0.18 

Granite 0.12 

Limestone   0. 10 

Sandstone 0.08 

Slate 0.36 

The  formula  may  be  expressed  as  follows : 

Rule. —  To  obtain  the  safe  uniformly  distributed  load  in 
tons  that  a  stone  lintel  will  support,  square  the  depth  of  the 
beam  in  inches  and  multiply  by  the  breadth  in  inches.  Divide 
this  product  by  the  span  of  the  lintel  in  inches;  then  multiply 
this  last  result  by  the  coefficient  given  in  the  table  for  the 
particular  stone  used. 

EXAMPLE. — A  limestone  lintel  is  20  inches  wide  and  14  inches  thick, 
spanning  an  opening  of  42  inches.  What  distributed  load  will  this 
beam  safely  carry  ? 

b  d"* 
SOLUTION. — The  safe  distributed  load  in  tons  =  —=—  X  coefficient. 

Then,  W  =  2°X  14X14  x  0. 10  =  9.38  tons, 

or  9.33  X  2,000  =  18,660  pounds.     Ans. 

If  the  load  is  concentrated  at  the  center  of  the  span,  the 
safe  load  will  be  one-half  the  safe  uniform  load  given  by  the 
above  rule. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  section  modulus  of  a  rectangular  section  10  inches 
wide  by  10  inches  deep  ?  Ans.  166f . 

2.  Calculate  the  section  modulus  of  a  10-inch  I  beam,  the  sectional 
area  of  which  is  11.8  square  inches.  Ans.  36.87. 

3.  What  must  be  the  width  of  a  yellow-pine  beam  to  support  a  uni- 
formly distributed  load  of  10,380  pounds,  the  span  of  the  beam  being 
18  feet,  its  depth  12  inches,  and  the  safety  factor  5  ?  Ans.  8  in. 

4.  A  uniformly  distributed  load  upon  a  steel  I  beam  having  a  span 
of  24  feet  is  25,000  pounds.     What  size  of  beam  will   be  the  most 
economical  to  use,  the  allowable  unit  stress  on  the  material  being 
18,000  pounds  ?  Ans.  15" ;  41.2  Ib. 

5.  It  is  desired  to  span  an  opening  20  feet  wide  in  an  18-inch  wall, 
laid  in  lime  mortar,  with  two  steel  channels  to  support  a  solid  brick 


ARCHITECTURAL  ENGINEERING. 


101 


wall  over  the  opening.    What  will  be  the  size  and  weight  of  the  channels 
if  a  safe  fiber  stress  of  15,000  pounds  is  adopted  ?        Ans.   12"  ;  19.8  Ib. 

6.  A  builder  wishes  to  determine  which  will  be  the  cheaper  :   to 
span  a  20-foot  opening  in  a  12-inch  brick  wall  laid  in  cement  mortar 
with  two  steel  channels,  or  with  yellow-pine  timber ;  the  steel  beam 
being  worth  If  c.  per  pound  and  the  yellow-pine  timber  being  worth 
§25  per  M.     The  load  to  be  supported  is  the  solid  brick  wall  above ;  a 
factor  of  safety  of  4  will  be  required  if  the  steel  I  beam  is  used ;  if  the 
wooden  beam  is  used,  a  factor  of  safety  of  6  will  be  adopted.     If  the 
builder  adopts  the    cheaper,   (a)  which  will  be  used  ?     (b)  how  much 
will  he  save  over  the  other  ?  ,         \  (a)  Yellow-pine  timber. 

'  1  (b}  §6. 10. 

7.  What  safe  uniformly  distributed  load  will  a  granite  lintel,  16 
inches  deep  and  24  inches  wide,  sustain,  its  span  being  6  feet  ? 

Ans.   10. 24  tons. 

8.  A  sandstone  beam  is  required  to  support  a  uniformly  distributed 
load  of  4  tons ;  it  is  necessary  that  it  should  be  18  inches  wide ;  what 
should  be  the  depth  of  this  beam,  if  the  span  is  8  feet  ?        Ans.   16i  in. 

9.  A  heavy  bluestone  flag,  6  inches  thick  and  5  feet  wide,   spans 
a  culvert  which  is  7  feet  wide.     What  safe  uniformly  distributed  load 
will  this  flag  support  ?  Ans.  4.6  tons. 


TRUSSED  BEAMS. 

114.  When  wooden  girders  of  great  span  are  heavily 
loaded,  it  becomes  necessary  to  strengthen  them  with  iron 
or  steel  camber  rods,  as  shown  in  Figs.  63  and  64.  Fig.  63 


FIG.  63, 


shows  a  girder  with  one  support  in  the  center,  and  Fig.  64 
shows  a  girder  with  two  supports.  The  span  of  the  beam 
or  girder  may  be  considered,  in  each  case,  as  the  distance 
between  the  supports,  the  strength  of  the  girder  being 
thereby  materially  increased. 


102 


ARCHITECTURAL  ENGINEERING. 


115.     Stresses  in  a  Beam  Witli  One  Strut. — In  Fig. 

Go,  let  UT represent  the  load  concentrated  at  D.     Then  the 


FIG.  64. 


stress  in  the  member  D  C  is  equal  to  W.  The  stress  in  the 
other  members  may  be  found  by  applying  the  following 

Kule. — I.  To  find  t/ic  stress  in  A  C  or  B  C,  divide  the 
length  of  line  A  C  by  the  length  of  the  line  I)  C,  and  then 
multiply  this  result  I)}'  one-half  of  the  load  IV. 

II.  To  find  the  stress  in  the  beam  A  B,  divide  the  length 
of  the  line  A  D  by  t/ie  length  of  the  line  D  C,  and  multiply  by 
one-half  of  the  load  W. 

In  the  diagram,  Fig.  65,  the  members  represented  by  solid 
lines  are  in  compression,  and  those  shown  dotted  are  in  tension. 

The  length  of  the  members  in  the  above  rules  may  be 
taken  in  feet  or  inches,  but  all  lengths  should  be  taken  in 


c 
FIG.  65. 

the  same  unit  of  measurement.     The  rules  may  be  expressed 
by  the  formulas 

Stress  in  D  C  =  +  W.  (2O.) 

ACor£C        -L£x~.         (21.) 


Stress  in  A  R  = 


(22.) 


ARCHITECTURAL  ENGINEERING. 


103 


The  -f-  and  —  signs  in  the  formulas  indicate  compression 
and  tension^  respectively.  The  -J-  sign  denotes  that  the 
result  obtained  is  a  compressive  stress.  The  —  sign  means 
that  the  result  is  a  tensile  stress. 

When  a  beam  with  one  support  at  center  has  a  uniform 
load,  as  in  Fig.  GG,  the  load  IV  on  the  center  strut  to  be  used 
in  formulas  £O  to  22  is  |  of  the  entire  load. 


FIG.  66. 

EXAMPLE. — What  is  (a)  the  tension  in  the  camber  rod,  and  (b)  the 
compression  on  the  trussed  beam  of  the  dimensions  and  loads  shown  in 
Fig.  66  ? 

SOLUTION. — We  must  first  compute  the  load  W  coming  upon  the 
strut  D  C.  The  load  is,  in  this  case,  usually  considered  equal  to  one- 
half  the  entire  load  on  the  beam.  But  as  the  beam  is  composed  of  one 
length  of  timber,  and  is  not  hinged  at  D,  being,  in  effect,  a  continuous 
beam,  the  load  on  the  center  strut  is  f  of  the  entire  load  on  the  beam. 
The  entire  load  on  the  beam  is  equal  to  30  X  1,000  =  80,000  pounds; 
|  of  30,000  =  18,750  pounds,  the  load  fF  acting  on  the  beam  directly 
over  the  strut  D  C. 

Then  (a)  the  tension  in  the  camber  rod  A  C  is  equal  to  the  length 
A  C-h  D  C  multiplied  by  \  of  W,  or,  substituting  the  given  dimensions, 
15.2H-2.5  =  6.08;  and  6.08  X1?  of  18,750  =  57,000  pounds,  the  tensile 
stress  on  rod  A  C.  Ans. 

(b)  To  determine  the  stress  on  beam  A  J?,  divide  the  length  A  D  by 
DC  and  multiply  by  \  of  W,  Thus  15-*- 2. 5  =  6;  6  X  J-  of  18,750 
=  56,250  pounds,  the  compressive  stress  in  the  beam  A  B.  Ans. 

116.  Stresses  in  Beams  Witli  Two  Struts. — In  Fig.  G7, 
the  calculations  for  the  stresses  in  the  various  members  are 
similar  to  those  given  for  the  trussed  beam  with  one  support. 
In  the  two- trussed  beams,  the  stress  in  B  //or  CR '  =  W. 
The  stresses  in  the  other  members  may  be  expressed  by 
rule,  as  follows: 


104  ARCHITECTURAL  ENGINEERING. 


Rule.— I.  To  obtain  the  stress  in  A  H  or  D  E,  divide  the 
length  of  A  H  by  the  length  of  B  H,  and  multiply  this  result 
by  the  load  \V. 

II.  To  find  the  stress  in  A  D  or  HE,  divide  the  length  of 
the  line  A  B  by  the  length  of  the  line  B  H,  and  multiply  this 
result  by  the  load  W. 


H 


FIG.  67. 


The  above  may  be  expressed  in  formulas: 

Stress  in  B  H  or  C  E  =  +  W. 

AH 


in  A  H  or  D  E  -. 

in  HE  =  - 


W. 


(23.) 
(2±.) 

(25.) 
>•) 


A  B 

Stress  in  A  D  =  +  -j^-fj  X  W. 
b  Jtl 

Compression  is,  as  previously  noted,  indicated  by  the 
+  sign  and  tension  by  the  —  sign. 

When  a  beam  has  two  supports  \  of  its  length  from  each 
end,  and  is  uniformly  loaded,  as  in  Fig.  68,  the  load  W  on 
each  support  is  \\  of  the  total  load. 

w 


-10OO  Ib  per  running  foot. 

' 


EXAMPLE. — A  beam  is  trussed,  as  shown  in  Fig.  68;  what  is  (a)  the 
stress  in  camber  rod  HE,  and  (b)  the  compression  in  the  beam  ADI 


§  5  ARCHITECTURAL  ENGINEERING.  105 

SOLUTION.— The  entire  load  on  the  beam  A  D  is  30  X  1,000  =  30,000 
pounds.  The  beam  being  a  continuous  girder,  as  in  the  previous 
example,  the  loads  //'are  each  \\  of  the  entire  load.  Hence,  //'  =  -J-J- 
X  30,000,  or  11,000  pounds. 

(a)     Applying  formula  25,  \vc  have 

Stress  in  HE  =  g-g  X  11,000  =  44,000  pounds.     Ans. 
(b~)     Applying  formula  20,  we  have 

Stress  in  .•/  D  =  ^  X  11,000  =  44,000  pounds.     Ans. 

r*l.  O 

117.  Graphical  Metliod  of  Computing-  Stress. — -The 
stresses  in  the  various  members  of  a  trussed  beam  may  be 
obtained  by  means  of  a  graphical  method  which  is  simply  an 
application  of  the  principles  of  the  resolution  of  forces.  (See 
Arts.  19  and  2O.)  Although  not  as  exact  in  its  results  as 
the  mathematical  method,  it  is  probably  more  satisfactory, 
there  being  less  chance  of  errors  creeping  into  the  calcula- 
tion. This  method  is  fully  explained  in  the  subjoined  illus- 
trative example : 

B (Separators  2*4  Wood. 


xlQ  Pott. 

24  Ft. Center  to  Center- 

Fio.  (iO. 


A  floor  is  to  be  supported  by  yellow-pine  girders,  each 
composed  of  two  4"  X  13"  beams,  trussed  with  a  wrought-iron 
rod,  as  shown  in  Fig.  69.  The  span  of  the  girders  is  34  feet, 
and  they  are  spaced  8  feet  from  center  to  center.  The  load 
is  light,  amounting  to  only  40  pounds  per  square  foot  of  floor 
surface.  Required,  to  determine  whether  the  two  yellow- 
pine  beams  are  sufficiently  strong,  and  what  should  be  the 
size  of  the  wrought-iron  camber  rod;  also  to  design  the 
detail  construction  for  the  parts  A  and  />. 

The  floor  area  supported  by  each  girder  is  34x8  =  11)3 
square  feet;  therefore,  the  total  load  on  a  girder  is  11)3x40 
=  7,680  pounds.  To  find  the  stress  produced  in  the  different 


106 


ARCHITECTURAL  ENGINEERING. 


members  of  the  truss  by  this  load,  first  draw  to  some  con- 
venient scale,  as  in  Fig.  70,  making  the  lines  ab,  ac,  be, 
and  dc  correspond,  respectively,  to  the  center  lines  of  the 
girder,  the  wro light-iron  camber  rod,  and  the  strut;  thus, 
the  line  a  b  represents  the  center  line  of  the  pine  beams,  its 
length  being  equal  to  24  feet  on  the  assumed  scale;  while 
dc,  drawn  perpendicular  to  ab  at  its  middle  point,  repre- 
sents, on  the  same  scale,  the  length,  20  inches,  of  the  strut. 


In  accordance  with  the  statements  of  Art.  115,  the  load 
carried  by  the  strut  may  be  taken  as  f  of  the  total  load  on 
the  girder  ;  therefore,  the  force  /",  Fig.  70,  acting  downwards 
on  the  frame,  and  borne  directly  by  the  strut  dc,  is  7,680 
Xf  =  4,800  pounds.  This  force  is  held  in  equilibrium  by 
the  stresses  in  the  members  of  the  truss,  represented  by  the 
center  lines  ad,  db,  ac,  and  c  b,  one-half  of  it,  or  2,400 
pounds,  being  held  by  each  of  the  pairs  a  d  and  a  c,  db  and 
c  d.  Considering  the  half  of  the  load  carried  by  the  pair  a  d 
and  ac,  we  have  a  downward  force  of  2;400  pounds,  and  it 
is  required  to  resolve  it  into  two  components,  one  acting 
along  the  line  a  c  and  the  other  along  ad.  Assuming  a 
scale  of  forces,  one,  for  example,  in  which  a  line  1  inch  long 
represents  a  force  of  800  pounds,  draw  the  line  dc,  Fig.  71, 
parallel  to  the  center  line  dc,  Fig.  70,  of  the  strut,  and 
make  its  length  correspond  to  a  force  of  2,400  pounds,  the 


|  inches   or  17200  Ib. 


FIG.  n. 


part  of  the  total  load  on  the  strut  which  is  borne  by  the 
members  ad  and  ac.  From  the  upper  extremity  of  dc, 
Fig.  71  draw  the  line  da  parallel  to  the  line  da  of  Fig.  70, 


§  5  ARCHITECTURAL  ENGINEERING.  107 

and  from  the  lower  extremity  draw  the  line  ca  parallel  to 
c  a  of  Fig".  70,  prolonging  these  two  lines  until  they  meet  in 
the  point  a.  The  lines  da  and  ca  of  Fig.  71  represent,  on 
the  scale  of  forces  to  which  the  line  dc  was  drawn,  the 
stresses  in  the  corresponding-  members  of  the  girder.  With 
the  assumed  scale  of  1  inch  =  800  pounds,  the  line  d ' c  must 
be  2,400-5-800  =  3  inches  long;  by  measurement,  the  lines 
da  and  c  a  are  found  to  be  2H  and  21f  inches  long,  respect- 
ively; therefore,  the  stress  represented  by  the  line  da  is 
2HX800  =  17,200  pounds,  and  that  represented  by  ca  is 
21f  X800  =  17,400  pounds.  The  stress  of  17,200  pounds  is 
the  total  compressive  stress  produced  in  the  two  yellow-pine 
beams  through  the  action  of  the  downward  thrust  on  the  strut. 

From  Table  6,  Art.  61,  it  will  be  found  that  the  ultimate 
resistance  to  compression  of  yellow  pine  per  square  inch  is 
4,400  pounds;  and  as  wood  is  not  so  reliable  as  iron,  it  is 
considered  advisable  to  use  a  factor  of  safety  of  G,  as  against 
a  factor  of  safety  of  4  for  the  camber  rods.  Since  the 
trussed  girder  is  secured  against  lateral  deflection  by  the 
floor  joist,  and  as  it  is  secured  from  deflection  in  a  vertical 
direction  at  the  center  by  the  load  upon  the  floor  and  by 
the  camber  rod  and  strut,  the  length  of  the  wooden  girder, 
which  may  be  considered  as  a  column  under  compressive 
stress,  is  only  one-half  the  span,  or  12  feet.  Now,  the  sec- 
tional dimension  of  the  girder  is  so  great  in  comparison 
with  its  length,  that  it  is  not  necessary  to  apply  the  column 
formula,  and  its  strength  may  be  considered  as  its  resistance 
to  direct  compression.  Hence,  4, 400  -r-G  =  733  pounds, 
which  is  the  allowable  compression  strength  of  the  girder 
per  square  inch  of  section.  Then,  17,200  pounds  (the 
compression)  -4-  733  pounds  (the  allowable  unit  stress)  =  23 
square  inches  required  to  take  care  of  the  compressive 
stress.  As  the  girder  is  known  to  be  12  inches  in  depth,  it 
is  readily  seen  that  this  compressive  stress  will  require  a 
section  of  the  timber  girder  equal  to  2  in.  X  12  in. 

There  is,  in  addition  to  this,  a  transverse  stress  upon  one- 
half  of  the  girder  produced  by  the  uniformly  distributed 
load.  To  find  the  amount  of  this  bending  stress,  consider 


108  ARCHITECTURAL  ENGINEERING.  §  5 

the  left-hand  half  of  the  girder  as  a  simple  beam,  sustaining  a 

uniformly  distributed  load  equal  to  one-half  of  the  total  load 

upon  the  girder,  that  is,  a  load  of  7, 680 -i- 2  =  3, 840  pounds. 

Applying  formula  13,  Art.  97,  the  bending  moment  due 

to  this  load  is  J/  =  -  -  =  5, 760    foot-pounds ;    and 

o 

5,760x12  =  69,120  inch-pounds.  Then  the  section  modu- 
lus required  may  be  obtained  by  the  formula  K  =  -~r-,  where 

K  equals  the  section  modulus,  Jlf  the  bending  moment  in 
inch-pounds,  and  ^  represents  the  allowable  unit  fiber  stress 
of  the  material,  which  is  equal,  in  this  case,  to  7,300  (the 
modulus  of  rupture  of  yellow  pine)  -4-  6  (the  factor  of  safety) 
=  1,217  pounds.  Substituting  the  values  in  the  above 

formula,  the  calculation  will  be  R  =       '         =  56.8,  the  sec- 

J_  j  £  J.  | 

tion  modulus  required  in  the  girder  to  successfully  resist  the 
transverse  stress. 

Since  the  section  modulus  of  a  rectangular  beam  may  be 

obtained  by  formula  15,  Art.  1O1,  which  is  K  =  — — ,  b 

being  the  width  of  the  beam  in  inches,  and  d  the  depth,  and 
as  K  is  already  known  to  be  56.8  and  the  depth  of  the  beam 
to  be  12  inches,  the  width  of  the  beam  required  to  resist  the 
transverse  stress  may  be  obtained  by  transposing  the  for- 
mula to  b  —  —-™-\  the  values  substituted  would  give 

56  8  V  6 
b  —  -        V»    or  2.37   inches,   which   is   the  width   of  the 

J..-V  /\  -L<4 

required  beam.  Then  adding  the  size  of  the  timber  required 
to  resist  compression  and  the  size  of  timber  required  to 
resist  the  transverse  stress,  we  have  a  timber  2  inches  wide 
by  12  inches  deep,  added  to  a  timber  2.37  or,  say,  2-£  inches 
by  12  inches,  which  gives  a  piece  44  inches  by  12  inches.  In  the 
girder  there  are  two  4"xl2"  timbers,  and  as  only  a  single 
4£/"X  12"  timber  is  required,  it  is  evident  the  girder  is  nearly 
twice  as  strong  as  is  necessary.  It  must,  however,  be  borne 
in  mind  that  the  theoretical  dimensions  of  members  do  not 
always  agree  with  those  required  in  practical  rules;  for 


ARCHITECTURAL  ENGINEERING. 


109 


Upset  to  1%  did  on  end. 


instance,  in  the  above  case  it  would  not  be  good  practice  to 
make  the  combined  sectional  area  of  the  girder  equivalent 
to  that  of  a  4 A "X  12"  timber,  as  obtained  by  the  calculation 
as  being  correct,  because  this  would  make  each  timber  a 
little  larger  than  2  in.  X  12  in.,  and  no  timber  or  girder, 
especially  where  rafter  or  flooring  is  spiked  to  it,  should  be 
less  than  3  inches  wide. 

From  Table  6,  the  ultimate  tensile  strength  of  wrought  iron 
is  found  to  be  50,000  pounds  per  square  inch ;  hence,  if  we  use 
a  factor  of  safety  of  4,  the    jrrt.iron  washer. 
safe  working  fiber  stress 
in  the  rod  must  be  50,000 
-=-4  =  12,500  pounds  per 
square  inch. 

According  to  the  results 
given  by  the  diagram,  the 
total  stress  in  the  rod  is  17, 400  pounds;  therefore,  the  rod  must 
have  a  section  of  17,400^12,500  =  1.39  square  inches.  The 
area  of  a  If -inch  round  rod  is  1.48  square  inches,  and  as  this  is 
the  nearest  standard  size  having  the  required  sectional  area, 
it  will  be  used.  As  the  area  at  the  bottom  of  the  thread  of  a 
If -inch  bolt  is,  however,  only  1.06  square  inches,  it  will  be 
necessary  to  upset  or  enlarge  the  ends  of  the  rod  to  a  diam- 
eter of  If  inches,  in  order  to  get  the  requisite  strength  in 
the  threaded  portion.  The  washer  at  B,  Fig.  69,  must  be  large 
enough  to  distribute  the  pressure  due  to  the  pull  of  the  rod 
over  a  sufficient  area  of  the  end  of  the  beams  to  prevent 

danger  of  crushing  the 


Ijtltottnd  Hod. 


FIG.  72. 


$  Bolts  with  wood  sepa  ra  tor. 


^  1  dia.dowelr  ^ 
cast  on. 


FIG.  73. 


square   inches,    nearly.      Using 


wood.  The  allowable 
compressive  strength 
of  yellow  pine,  par- 
allel to  the  grain, 
may  be  taken  as  800 
pounds  per  square 
inch;  this  requires  a 
washer  whose  area  is 
17,400-4-800  =  22 
a  washer  6  inches  wide, 


110  ARCHITECTURAL  ENGINEERING.  §  5 

extending  across  the  ends  of  the  two  beams,  we  get  a  bear- 
ing area  of  2  X  4  X  G  =  48  square  inches.  In  order  to  resist 
the  bending  stress  due  to  the  pull  of  the  rod,  the  washer 
should  be  from  f  inch  to  1  inch  in  thickness. 

Figs.  G!J,  72,  and  73,  which  are  so  clearly  drawn  as  to 
require  no  further  explanation,  show  excellent  details  for  the 
different  parts  of  the  trussed  stringer  under  consideration. 


DEFLECTION    OF    FLOORBEAMS. 

118.  Beams  used  in  floors  should  not  only  be  strong 
enough  to  carry  the  siiperimposed  loads,  but  also  sufficiently 
rigid  to  prevent  vibration.  For  beams  carrying  plastered 
ceilings,  if  the  deflection  exceeds  ^T  of  the  distance 
between  the  supports,  or  ^  of  an  inch  per  foot  of  span, 
there  is  danger  of  cracking  the  plaster.  It  is  safe  to  assume 
that  these  deflections  will  not  be  exceeded  in  wood  beams, 
and  the  sagging  of  the  beam  will  not  produce  plaster 
cracks  when  the  beam  is  loaded  with  its  maximum  safe 
load,  if  the  depth- of  the  beam  is  made  Jy  of  the  span.  That 
is,  suppose  the  span  of  the  beam  is  20  feet,  or  240  inches: 
TV  of  240  in.  =  13.3,  say,  14  in.,  the  depth  of  the  wood 
beam  to  be  used  for  this  span.  If  the  calculation  brings  out 
a  result  with  a  decimal,  like  that  just  given,  take  the  next 
stock  size  above  the  result  thus  obtained. 

If  steel  beams  have  a  depth  in  inches  equal  to  one-half  of 
the  span  in  feet,  their  deflection  may  be  considered  well 
within  the  safe  limit.  Thus,  if  the  span  of  a  steel  beam  is 
24  feet,  its  depth  should  not  be  less  than  12  inches. 

It  must  be  remembered  that  the  above  rules  are  only 
approximate,  and  if  there  is  any  reason  to  suppose  that  the 
deflection  of  the  beam  will  be  excessive,  its  deflection  should 
be  computed  by  the  rules  and  formulas  subsequently  given 
in  the  second  section  of  this  subject.  The  above  rules  may, 
however,  be  considered  comparatively  safe,  and,  if  followed, 
are  not  likely  to  produce  plaster  cracks,  provided  the  loads 
are  statical  and  the  floors  not  subject  to  sudden  shocks  or  jars. 


§5  ARCHITECTURAL  ENGINEERING.  Ill 

EXAMPLES   FOR   PRACTICE. 

1.  It  is  found  necessary  to  truss  the  yellow-pine  purlins  supporting 
a  roof,  with  a  wrought-iron  camber  rod  on  each  side  of  the  purlin ;  the 
length  of  the  purlin  is  20  feet,  the  depth  of  the  truss  from  the  center  of 
the  rods  to  the  center  of  the  purlin  is  14  inches,  and  the  load  upon  the 
central  strut   is  1,600  pounds.     What  should  be  the  diameter  of  the 
camber  rods  if  the  ends  of  the  rods  are  upset,  and  a  safety  factor  of  4 
is  desired  ?  Ans.   |  in.  diam. 

2.  A  girder  of  24  feet  span  is  trussed  at  the  center  by  a  camber  rod 
and  strut ;  the  depth  of  the  truss  from  the  center  of  the  girder  to  the 
center  of  the  rod  is  2  feet ;   if  the  beam  is  loaded  with  a  uniformly 
distributed  load  of  2,000  pounds  per  lineal  foot,   (<?)  what  is  the  stress 
on  the  rod  ?     (b)  what  is  the  compressive  stress  on  the  beam  ?     (c)  what 
is  the  stress  on  the  central  strut  ?  ,    (a)  91,200  Ib. 

Ans.  -    (b)  90,000  Ib. 
(  (c)  30,000  Ib. 

GRAPHICAL  STATICS. 

119.  The  application  of  the  principles  of  the  triangle  of 
forces,  in  finding  the  stresses  in  the  different  members  of  a 
simple  trussed  stringer,  have  been  illustrated  in  Art.  117. 
We  will  now  show  how  this  process  may  be  extended,  by  the 
use  of  the  method  of  the  polygon  of  forces  (see  Art.  17),  so 
as  to  make  it  available  in  computing  the  stresses  in  the  mem- 
bers of  the  most  complicated  frames  and  trusses  found  in 
ordinary  engineering  practice.  The  application  of  the  poly- 
gon of  forces  to  the  computation  of  stresses  is  called  graph- 
ical statics. 

In  the  use  of  graphical  statics  to  determine  the  stresses  on 
the  various  members  entering  into  the  construction  of  a 
frame,  not  only  may  the  magnitude  of  the  stresses  upon  the 
members,  but  the  direction  in  which  they  act,  be  determined. 

The  representation  to  the  eye  of  the  forces  existing  in  the 
several  parts  of  a  frame  structure,  possesses  many  advantages 
over  their  determination  by  calculation.  Graphical  analysis 
being  founded  on  correct  principles,  the  diagrams  give 
results  depending  for  accuracy  on  the  exactness  with  which 
the  lines  have  been  drawn,  and  upon  the  scale  by  which  they 
are  measured.  With  ordinary  care,  the  different  forces  may 


ARCHITECTURAL  ENGINEERING. 


1000  II. 


be  obtained  much  more  accurately  than  the  several  parts  of 
the  frame  can  be  proportioned. 

12O.     Frame  and  Stress  Diagrams. — In  Fig.  74,   the 

weight    W  of  1,000  pounds  is  supported  at  c  by  the   two 

branching  cords  c  a  and  c  b,  fastened  to  the  pins  a  and  b.    The 

figure,  which  is  drawn  to  scale  and  accurately  represents  the 

outline  of  the  structure,  is 
called  a  frame  diagram. 
Now,  to  obtain  the  stress 
upon  the  cords  c  a  and  c  b, 
draw  the  vertical  line  1-2  in 
Fig.  75;  this  line  must 
represent  the  direct  pull  on 
the  rope,  or  cord,  c  W,  so 
let  each  inch  represent  400 
pounds;  then,  to  represent 
a  force  of  1;000  pounds,  the 

length  of  the    line   1-2  shall   be    1,000 -=-400  =  2|   inches. 

From  1  draw  the  line  1-3  parallel  to  a  c,  and  from  2  draw 

the  line  2-3  parallel  to  the  line  c  b  1 

of  Fig.  74;  they  "will  intersect  at 

the  point  3  and  form  a  triangle. 

If  the  lines  1-3  and  2-3  are  meas- 
ured  with   the   1-inch    scale,    the 

stresses  c  a  and  c  b  can  be  deter- 
mined. 

The  diagram,  Fig.  75,  is  called 

the  stress  diagram.     In  working 

out  the  stresses  for  roof  truss,  it  is 

first  necessary  to  make  the  frame 

diagram,  drawn  accurately  to  any 

scale  and  representing  the  outline 

of  the  truss  and  the  members  of 

which  it  is. composed.      It  is  then 

necessary  to  draw  a  stress  diagram 

for  the  dead  load  on  the  roof,  which  FIG.  75. 

sometimes  includes  the  snow  load,  and  another  diagram  to 


ARCHITECTURAL  ENGINEERING. 


113 


represent  the  stresses  produced  by  the  action  of  the  wind  on 
the  roof. 


Lettering  the  Diagrams. — In  laying  out  the 
frame  and  stress  diagram,  it  is  useful  to  adopt  a  system  of 
lettering,  so  that  the  relative  position  of  the  different  mem- 
bers in  the  frame  and  stress  diagram  may  be  seen  at  a  glance, 
as  in  Figs.  76  and  77.  In  Fig.  70,  the  reactions  at  the  walls 
are  /^  and  Ra. 

It  must  be  remembered  that  a  roof  truss  is  nothing  more 
nor  less  than  a  beam,  and,  consequently,  the  sum  of  the  reac- 
tions must  be  equal  to  the  sum  of  the  loads. 

The  loads,  or  forces,  acting  upon  a  roof  truss  are  always 
considered  as  concentrated  at  the  panel  points,  that  is, 
where  several  members  in  the  structure  join  each  other.  In 


Fig.  76,  the  loads  and  reactions  on  the  truss  are  shown  by 
dotted  lines  and  arrowheads. 

The  spaces  outside  of  the  truss,  between  the  loads,  together 
with  each  triangle  inside  of  the  truss,  should  be  lettered  with 
capitals,  as  in  Fig.  76.  It  is  well  to  begin  at  the  left-hand 
reaction  of  the  truss  with  the  letter  A,  working  around  the 
outside  of  the  truss  in  alphabetical  order,  until  the  right-hand 
reaction  is  reached.  Then  start  with  the  first  triangular 
space  at  the  left-hand  end  of  the  truss,  following  with  the 
next  letter  in  the  alphabet,  continuing  in  alphabetical  order, 


114 


ARCHITECTURAL  ENGINEERING. 


it  being  well  to  mark  the  space  between  the  two  reactions,  at 
the  center  of  the  truss,  Z. 

It  is  not  absolutely  necessary  to  use  this  system  of  letter- 
ing, for  any  letters  or  numbers  may  be  employed,  as  long  as 
no  two  spaces  in  the  truss  are  marked  alike.  It  is,  however, 
well  to  have  some  definite  system  in  engineering  work,  the 
above  being  as  convenient  as  any  that  can  be  suggested. 

Having  marked  the  truss  with  letters,  as  in  Fig.  76,  the 
various  members  and  forces  may  be  designated  by  those 
letters  between  which  they  are  located.  The  left-hand  reac- 
tion, for  instance,  will  be  ZA,  the  first  load  on  the  truss  A  B, 
the  second  load  B  C,  the  load  at  the  apex  of  the  truss  CD, 

a  next  in  succession  the 
loads  DE  and  EF — al- 
ways following  the  truss 
in  the  same  direction. 
The  last  external  force 
on  the  truss  is  the  right- 
hand  reaction  F  Z, 

The  lower  portion  of 
the  left-hand  rafter  is 
B  G,  the  upper  portion 


C  //,  the  left-hand  por- 
1  a  tion  of  the  tie  at  the  base 
of  the  truss  is  G  Z,  the 
left-hand  compression 
member  in  the  truss  is 
HG,  and  the  central  tie 
///,  and  so  on.  In  desig- 
nating the  various  mem- 
/  bers  of  the  truss,  the  let- 
ters are  given  in  the  order 
in  which  they  occur,  always  following  the  joints  in  the  same  di- 
rection, which  in  this  case  is  that  in  which  the  hands  of  a  watch 
travel.  This  being  important,  it  is  fully  explained  further  on. 


FIG.  77. 


The  stress  diagram  for  the  truss  shown  in  Fig.  76, 
is  represented  by  Fig.   77.     The  stress  diagram  is  always 


ARCHITECTURAL  ENGINEERING. 


115 


(Fig. 


drawn  to  some  scale  in  which  the  unit  measurement  repre- 
sents a  certain  number  of  pounds.  Thus,  suppose  1  inch 
represents  1,000  pounds,  and  suppose  that  in  the  stress  dia- 
gram the  line  c Ji  is  -i  inches  long;  then  the  stress  in  the 
member  C H  in  Fig.  TO,  is  equal  to  4,000  pounds.  It  will  be 
noticed  that  the  small  letters  are  used  in  the  stress  diagram 
and  that  the  letters  at  the  end  of  a  line  designate  that  line. 
Thus,  the  length  of  the  line  d i  in  the  stress  diagram.  Fig. 
77,  measures  the  stress  in  the  corresponding  member  /J/of 
the  frame  diagram,  Fig.  TO.  The  distance  from  z  to  a 
measures  the  magnitude  of  the  reaction  A  Z;  the  length  of 
the  line  ///  in  the  stress  diagram  measures  the  stress  in  the 
corresponding  member  ///of  the  frame  diagram. 

123.  Order  of  Letters. — It  is,  as  previously  men- 
tioned, very  important  always  to  read  the  letters  in  the 
same  direction,  and  in  proper  order.  If,  for  instance,  the 
joint  at  the  middle  of  the  left-hand  rafter 
examined,  the  members  and 
stresses  must  be  read  off  in 
their  proper  order,  as  B  C, 
C  H,  H  G,  and  back  again  to 
G  B.  Leaps  should  not  be 
made  from  B  C  to  G  />,  etc. , 
as  this  would  lead  to  errors, 
and  prevent  the  drawing  of 
the  stress  diagram.  Still 
more  important  is  it  to  read 
around  the  joints  in  one  direction,  as  in  Fig.  78,  i.  e.,  in  the 
direction  of  the  arrow.  If  you  reverse  the  reading  of  the 
pieces,  you  must  reverse  the  direction  of  the  stresses  in  the 
diagram.  If,  for  instance,  in  Fig.  70,  G  H  was  read,  and  its 
corresponding  line  gh  found  in  the  stress  diagram,  its  direc- 
tion would  be  downwards,  pulling  away  from  the  joint  at  the 
left-hand  rafter  and  making  G  H  a.  tie-rod,  whereas  it  is  well 
known  as  a  strut.  If,  however,  it  had  been  read  correctly 
hg,  it  would  indicate  a  push  towards  the  joint,  which  is,  of 
course,  the  correct  action  of  a  strut. 

1-20 


FIG.  78. 


110 


ARCHITECTURAL  ENGINEERING. 


When  the  joint  Z,  Fig-.  76,  is  examined,  the  reverse  of  this 
occurs,  and  here  the  correct  reading  is  gli,  which  is  the  same 


relative  direction,  for  the  point  Z,  as  was  Jig  for  the  point 
at  the  center  of  the  left-hand  rafter. 

124.     Diagram   for  Simple  Frames. — Fig.  79  shows  a 

force,  or  load,  of  1 , 000  pounds 
pulling  upon  the  cord  EA, 
and  held  in  position  by  the 
two  cords  A  C  and  C E. 
Find  by  graphical  statics  the 
stress  in  these  two  cords, 
also  the  magnitudes  of  the 
forces  (A  B,  BC}(CD,DE] 
required  to  act  at  the  ends 
of  the  cords. 

Draw  the  frame  diagram, 
Fig.  79,  accurately,  say  to  a 
scale    of  f  inch    to   1    foot. 
Then  start  to  draw  the  stress 
diagram,  Fig.  80.     The  force 
J(t  EA,-in  the  frame  diagram, 
being  already  known,   take 
some  scale,  say  400  pounds  to  1  inch,  and  draw  ea  in  the 


FIG. 


§  5  ARCHITECTURAL  ENGINEERING.  11? 

stress  diagram.  It  must  be  drawn  parallel  to  the  line  along 
whieh  the  force  or  load  E  A  acts.  This  force  E  A  being 
1,000  pounds,  and  the  scale  to  which  the  stress  diagram  is 
drawn  being  -400  pounds  to  each  1  inch,  the  line  c  a  must  be 
2$  inches  long.  Having  drawn  c  a,  work  around  the  joint 
in  the  direction  of  the  arrow.  From  a  in  the  stress  diagram 
draw  a  line  parallel  to  A  C.  and  from  c  in  the  stress  dia- 
gram draw  a  line  parallel  to  C  E  until  it  intersects  the  line 
a  c  at  the  point  c. 

In  going  around  the  stress  diagram,  the  same  direction  is 
to  be  followed.  Thus,  in  going  around  the  joint  A  E,  it  is 
read,  in  the  stress  diagram,  from  e  to  a,  from  a  to  c,  from  c 
to  ^,  always  arriving  at  the  same  point  from  which  the  start 
was  made.  As  the  stresses  are  read,  their  direction  should 
be  marked  with  arrows  on  the  frame  diagram,  Fig.  79.  This 
shows  the  direction  of  the  stress  and  designates  whether  it  is 
compression  or  tension.  Forces  acting  away  from  a  joint  are 
always  tensile  stresses;  those  acting  towards  a  joint  are 
always  compressive  stresses. 

If  there  is  tension  at  one  end  of  a  member,  it  is  evident 
there  must  be  an  equal  amount  of  tension  at  the  other  end ; 
if  there  is  compression  at  one  end  of  a  member,  there  is  an 
equal  amount  at  the  other. 

An  easy  way  to  remember  whether  the  arrows  designate 
compression  or  tension  by  their  direction,  as  shown  on  the 
members  in  a  frame  diagram,  may  be  seen  by  referring  to 
Figs.  81  and  82.  The  member  A  B,  Fig.  81,  is  a  tension 
member;  the  arrows  point  a\vay  from  the  joints  and  towards 
each  other,  and  resemble  the  form  of  an  elastic  material, 
stretched,  as  shown  at  a,  Fig.  82 ;  while  in  the  compression 
member  A  C,  the  arrows  act  against  the  joints,  or  away  from 
each  other,  and  resemble  the  form  that  a  plastic  material 
assumes  on  being  compressed,  as  shown  at  d,  Fig.  82. 

To  continue  the  solution  of  the  problem  in  Fig.  79 :  Having 
gone  around  the  joint  E  A  C,  proceed  to  go  around  the  joint 
ABC  in'  the  same  direction.  Having  the  point  a,  draw 
from  it  a  line  parallel  to  A  It  in  the  frame  diagram,  then 
draw  a  line  from  c  parallel  to  the  line  B  C  in  the  frame 


118  ARCHITECTURAL  ENGINEERING.  §  5 

diagram ;  the  point  where  the  two  lines  intersect  is  /;.  The 
polygon  of  forces  may  be  read  from  c  to  a,  from  a  to  b,  from 
b  to  c,  always  moving  in  the  same  direction  and  arriving  at 
the  starting  point.  The  next  joint  to  work  around  is  the 


FIG.  81. 

joint  C  D  E.  Starting  at  the  point  c  already  determined, 
draw  a  line  parallel  to  the  direction  of  CD,  and  from  c  draw 
a  line  parallel  to  D  E ;  the  polygon  of  forces  is  from  a  to  c, 
c  to  d,  and  d  to  c,  the  last  point  in  the  diagram,  and  the 
point  from  which  the  whole  stress  diagram  was  started. 
Then  by  measuring  the  lines  (with  the  1-inch  scale,  wherein 
every  inch  represents  400  pounds)  in  the  stress  diagram,  the 


FIG.  82. 

stresses  in  the  different  members  of  the  frame  diagram  may 
be  found.  If,  for  instance,  it  is  desired  to  obtain  the  stress 
in  the  cords  A  C  and  C  E,  measure  the  lines  a  c  and  c  c  in 
the  stress  diagram ;  likewise,  to  obtain  the  forces  A  B,  B  C, 
C  D,  and  D  E,  measure  the  length  of  the  lines  a  b,  b  c,  c  d, 
and  d  c  in  the  stress  diagram. 

125.     In  Fig.  83  is  shown  a  case  similar  to  that  in  the 
preceding  article.     The  compression  member  B  D,  and  the 


ARCHITECTURAL  ENGINEERING. 


119 


tension    members  A  D  and  D  C  form    a  triangular   frame 
which  supports  the   downward   pull  of  1,000  potmds.      The 


FIG.  83. 

triangular  frame  is  supported,  in  turn,  by  the  reactions 
A  B  and  B  C,  Draw  the  stress  diagram  to  determine  the 
stress  in  the  various  members. 

Take  a  scale,  in  this  case  100  pounds  to  \  inch,  and  draw  the 
vertical  line  c  a,  Fig.  84,  equal  to  1,000  pounds.  This  line 
represents  the  force  C  A  in 
the  frame  diagram,  Fig.  83. 
Start  to  work  around  the 
joint  A  D  C,  in  the  direction 
of  the  arrow.  The  first 
member  encountered  is  A  D. 
Hence,  from  a  in  the  stress 
diagram  draw  a  line  parallel 
to  A  D  in  the  frame  dia- 
gram. Then,  D  C  being  the 
next  member  met  with,  from 
c  in  the  stress  diagram  draw 
a  line  parallel  to  D  C.  The 
point  of  intersection  of  the 
two  lines  just  drawn  is  d.  scale.  100  ib.  to  i  inch 
This  done,  go  around  the 
joint  again,  to  see  that  FIG-  84. 

none  of  the  members  have  been  omitted,  and  also  to  get 
the  direction  in  which  the  stresses  act.  Starting  at 


120  ARCHITECTURAL  ENGINEERING.  §  5 

c  in  the  stress  diagram,  and  going  around  the  joint  C  A  D, 
the  polygon  of  forces  is  as  follows :  from  c  to  a,  from  a  to  d, 
and  from  d  back  again  to  c,  thus  arriving  at  the  point  from 
which  the  start  was  made.  The  next  joint  in  the  frame 
diagram  is  A  B  D.  The  point  b  on  the  line  ca  is  not 
known,  but  may  be  determined  by  calculating  the  reactions 
A  B  and  B  C  in  the  same  manner  as  for  a  beam.  Thus,  the 
load  of  1,000  pounds  is  placed  upon  the  assumed  beam,  6  feet 
from  the  reaction  B  C.  The  moment  about  CB  is  1,000x6 
=  6,000  foot-pounds,  while  the  reaction  at  A  B  equals  6,000 


FIG.  85. 


-T-  13  =  461  pounds.  Knowing  that  the  force  A  B  is  461 
pounds  and  that  it  acts  upwards,  the  point  b  can  easily  be 
located  by  measuring  from  a  on  the  line  ac  in  the  stress 
diagram ;  then  the  line  b  d  may  be  drawn  and  if  found 
parallel  to  the  member  B  D  in  the  frame  diagram,  the  stress 
diagram  is  correct.  In  this  case,  however,  it  is  not  neces- 
sary to  calculate  the  reactions  A  B  and  B  C,  the  point  d 
having  been  already  determined,  and  since  we  know  that 
the  line  db  must  be  parallel  to  D  B,  all  that  is  needed  to 
locate  the  point  b  is  to  draw  a  line  from  d  parallel  to  D  B, 
and  the  point  where  it  cuts  the  line  c  a  is  b.  Having  found 
the  point  b  and  drawn  the  line  db,  go  around  the  joints 


ARCHITECTURAL  ENGINEERING. 


121 


A  B  D  and  C  D  />,  marking  the  direction  of  the  stress  by  the 
arrowheads,  as  shown  in  the  frame  diagram,  Fig.  83. 
Around  the  joint  A  B  D  the  polygon  of  forces  is  from  a  to 
b,  from  /;  to  d,  and  from  d  back  again  to  a.  Working 
around  the  joint  C  D  B,  the  polygon  of  forces  is  from  c  to  d, 
from  d  to  /;,  and  from  b  back  again  to  r.  This  completes 
the  stress  diagram ;  the  magnitude  of  the  stresses  in  the 
several  members  of  the  frame  diagram  is  found  by  measur- 
ing the  corresponding  lines  in  the  stress  diagram. 

126.  Diagram  for  a  Small 
Hoof  Truss. — Fig.  85  is  the  frame 
diagram  for  a  small  roof  truss. 
The  two  rafter  members  E  B  and 
C £  are  connected  at  their  foot  by 
the  tension  member  E  Z.  The 
loads  and  their  reactions  are  as 
shown  in  the  frame  diagram.  De- 
termine the  stresses  in  the  several 
members  composing  the  truss. 

Draw  the  vertical  line  a  d,  shown 
in  the  stress  diagram,  Fig.  80.  Lay 
off  to  any  scale,  say,  in  this  case 
2,000  pounds  to  £  inch,  the  load 
a  b  •  then,  to  the  same  scale,  the 
loads  be  and  c  d.  From  the  point 
d,  the  reaction  d  z,  8,000  pounds, 
acts  upwards,  which  determines 
the  point  s.  Now  go  around  the 
joint  A  B  E  Z.  The  reaction  Z  A 
acts  upwards  and  A  B  downwards. 
Then,  from  the  point  b  in  the  stress 
diagram  draw  the  line  be  parallel 
to  BE  in  the  frame  diagram,  and 
from  z  draw  the  line  e  z  parallel  to  FIG.  86. 

the  member  E  Z  in  the  frame  dia- 
gram.    The  point  of  intersection  will  be  the  point  c.    Having 
gone  thus  far,  again  go  around  the  joint,  to  get  the  direction  of 


Scale: 20OO  Ib.toi  inch. 


122 


ARCHITECTURAL  ENGINEERING. 


the  stress  in  the  members  and  to  see  whether  the  polygon 
of  forces  is  correctly  drawn.  Go,  for  instance,  from  z  to  a 
upwards;  a  to  b  downwards;  then  from  b  to  c,  and  from  e 
back  again  to  j,  the  starting  point.  The  next  joint  in  the  frame 
diagram  is  B  C E.  The  force  be  being  already  determined, 
from  point  c  draw  the  line  c  e,  parallel  to  the  member  C  E  in 
the  frame  diagram ;  this  line  passes  through  the  point  e  if  the 
diagram  has  been  drawn  correctly.  The  polygon  of  forces 
at  E  B  C  is  from  b  to  c  downwards,  then  from  c  to  e,  and  back 
again  from  c  to  b,  the  starting  point.  The  next  joint  in  the 
frame  diagram  is  E  C  D  Z.  and  the  polygon  of  forces  in  the 
stress  diagram  is  from  c  to  c  already  drawn,  from  c  to  d, 
d  to  z,  and  then  from  z  back  to  e..  * 

The  stress  diagram  completed,  all  that  remains  is  to  meas- 
ure the  various  lines  in  the  stress  diagram  which  represent 


FIG.  87. 

the  corresponding  members  in  the  frame  diagram.  Thus, 
cb  measures  If  inches,  the  scale  being  2,000  pounds  to  % 
inch;  hence,  the  stress  in  this  member  is  7,000  pounds; 
the  line  cs  measures  about  If  inches,  and  the  stress  in  the 
member  cz  is  5,500  pounds.  In  this  manner,  the  stress  in 
any  member  may  be  determined. 

127.     Diagram  for  a  Jib  Crane. — A  jib  crane  propor- 
tioned as  in  Fig.  87  has  a  load  of  30,000  pounds  suspended 


ARCHITECTURAL  ENGINEERING. 


123 


at  the  end  of  the  jib ;  what  are  the  stresses  in  the  guy  ropes 
and  in  the  different  members  of  the  crane,  and  what  are  the 
reactions  C A  and  DAI 

In  the  stress  diagram,  Fig.  88,  draw  the  vertical  line  be 
equal  to  30,000  pounds,  and  from  the  point  c  draw  the  line  c  c 
parallel  to  C  E  in  the  frame  diagram,  Fig.  87.  Then  from  b 
draw  the  line  eb  parallel  to  EB  in  the  frame  diagram. 
Again  going  around  the  joint  to  check  the  polygon  of  forces, 
they  are  found  to  be  from  c  to  e, 
from  e  to  d,  and  from  b  back 
again  to  c.  The  next  joint 
encountered  is  E  D  B.  Hence, 
from  e  draw  ed  upwards  par- 
allel to  E  D,  and  from  b  draw 
db  parallel  to  D  B,  the  point 
where  these  two  lines  intersect 
being  d.  The  polygon  of  forces 
about  the  joint  E  D  B  is  from 
b  to  ^,  from  e  to  d,  and  from  d 
back  again  to  b,  the  starting 
point.  Next,  go  around  the 
joint  C  A  D  E,  and  draw  ca 
upwards ;  then  from  d  draw  a  d 
parallel  to  A  D  in  the  frame 
diagram;  where  the  lines  just 
drawn  intersect  will  be  the 
point  a ;  de  and  e  c  have  already 
been  drawn.  The  remaining 
joint  to  work  around  is  A  B  D. 
On  looking  at  the  stress  dia- 
gram, it  may  be  seen  that  the  forces  around  this  joint  have 
already  been  determined,  completing  the  stress  diagram. 

The  stresses  in  the  members  may  be  determined,  as 
already  stated,  by  measuring  the  lines  corresponding  to 
them  in  the  stress  diagram,  with  the  scale  to  which  the 
diagram  .has  been  drawn. 

128.      Roof  Truss   With    a   40-Foot   Span.— Fig.    89 

shows  the  frame  diagram  for  a  40-foot  span  roof  truss.     The 


to-^-inch. 


124  ARCHITECTURAL  ENGINEERING.  §  5 

loads  are  as  shown,  the  compression  members  being  indi- 
cated by  heavy  lines,  and  the  tension  members  by  a  light 
line.  Required,  to  draw  the  stress  diagram  for  this  truss. 


FIG. 


First  draw  the  vertical  line  af  as  shown  in  the  stress 
diagram,  Fig.  90;  mark  the  point  a  and  lay  off  on  this  ver- 
tical line,  to  any  scale,  using,  in  this  case,  4,000  pounds  to 
|  inch,  the  loads  ab,  be,  c d,  dc,  and  ef,  corresponding  to 
the  loads  A  B,  B  C,  CD,  D  £,  and  E  F  in  the  frame  dia- 
gram. The  truss  being  symmetrically  loaded,  the  loads  are 
the  same  in  amount  on  the  two  sides  of  the  center  line. 
The  reactions  R^  and  R^  are,  therefore,  each  equal  to  one- 
half  of  the  load,  in  this  case,  16,500  pounds.  Hence,  za 
may  be  laid  off  on  the  vertical  line,  and  as  Rl  equals  Rt,  2  a 
must  equal  fs ;  consequently,  s  is  located  centrally  between 
a  and  _/",  or  between  c  and  d.  The  point  z  having  been 
determined,  proceed  with  the  diagram  by  going  around  the 
joint  A  B  G  Z.  Draw  b g  in  the  stress  diagram  parallel  to 
B  G  in  the  frame  diagram;  then  from  z  draw  gz  parallel  to 
G  Z,  the  point  where  the  two  lines  intersect  being  g.  The 
next  joint  is  B  C  H  G.  As  b  e  in  the  stress  diagram  is 
already  known,  draw  ch  parallel  to  C  //and  //£•  parallel  to 
H  G.  Then  the  polygon  of  forces  around  this  joint  will  be 


ARCHITECTURAL  ENGINEERING. 


125 


from  b  to  c,  from  c  to  it,  from  h  to  g,  and  from  g  back  again 
to  b.  It  is  now  expedient  to  analyze  the  joint  C  D I  H.  In 
the  stress  diagram,  h  c  and  c  d  have  already  been  obtained ; 
then,  from  d,  draw  di  parallel  to  D  I,  and  from  the  point  /t, 
already  known,  draw  Hi  parallel  to  I  H.  The  polygon  of 


Scale:  4000  Ib    to 


FIG.  90. 

forces  around  this  joint  will  be  from  c  to  d,  from  d  to  z,  from 
i  to_  //,  and  //  to  c,  the  starting  point,  arrowheads  always 
marking  the  direction  in  which  the  stresses  act  in  the  stress 
diagram,  upon  the  members  in  the  frame  diagram.  Now 
analyze  the  forces  around  the  joint  G  H IJ Z  in  which  the 
forces  zg,  gh,  and  h  i  have  been  obtained,  From  the  point 


120 


ARCHITECTURAL  ENGINEERING. 


z,  ij  is  drawn  parallel  to  //;  and  from  z,  j  z  is  drawn  parallel 
to  J  Z;  the  point  j  is  found  to  fall  on  the  pointy,  and  the 
polygon  of  forces  around  this  joint  is  from  z  to  g,  from  g  to  h, 
from  h  to  /,  from  i  to  j,  and  from/  back  again-  to  2,  the 
starting  point.  The  stress  in  the  members  around  the  joint 
ID EJ  should  next  be  determined,  j  i,  id,  and  de  being 
already  known.  From  c  draw  ej  parallel  to  EJ.  The  poly- 
gon of  forces  around  this  joint  will  then  be  from  i  to  d,  from 
d  to  i\  from  c  to/,  and  from/  back  again  to  i. 

The  only  remaining  joint  to  go  around  \sJEFZ.  By 
referring  to  the  stress  diagram,  it  is  seen  that  the  stresses 
in  these  members  have  been  determined,  while  the  polygon 
of  forces  around  this  joint  is  from  /  to  e,  from  e  to  f,  from 
/to  xr,  and  back  again  from  z  to/. 

The  stress  diagram  completed,  the  magnitudes  of  the 
stresses  may  be  determined  by  measuring  the  various  lines 
with  the  scale  to  which  the  diagram  has  been  drawn,  in 
this  case  4,000  pounds  to  £  of  an  inch. 


T 


FIG.  91. 


129.     Truss  for  a  Church  Roof. — Fig.  91  is  the  frame 
diagram  of  a  form  of  truss  sometimes  used  to  support  a 


§  5  ARCHITECTURAL  ENGINEERING. 


12', 


church  roof.  Determine  the  stress  diagram  for  the  dead 
load  and  also  the  stress  diagram  for  the  wind  pressure  on 
this  roof. 

First  draw  the  stress  diagram  (Fig.  92)  for  the  dead  load. 
As  the  dead  loads  upon  the  truss  are  symmetrical  both  in 


Scale:  2000  tb'  to  rj  inch. 
FIG.  92. 

amount  and  location  with  regard  to  the  center  line  of  the 
truss,  the  reactions  are  the  same  at  either  end  of  the  truss, 
and  each  one  is  equal  in  amount  to  one-half  of  the  load,  in 
this  case  7,575  pounds. 

Draw  the  vertical  line  af  in  the    stress  diagram ;  then, 
starting  at  the  point  a,  lay  off  the  scale  of,  say,  2,000  pounds 


128  ARCHITECTURAL  ENGINEERING.  §  5 

to  every  %  inch,  the  force  a  b  equal  to  A  B  in  the  frame  dia- 
gram; then  lay  off  be  equal  to  R  C,  cd  equal  to  CD,  de 
equal  to  D  E,  and  cf  equal  to  R  F.  Then,  as  the  truss  is 
symmetrically  loaded,  the  point  x  is  located  midway  between 
the  points  a  and  /.  If  the  truss  was  not  symmetrically 
loaded,  the  reactions  would  have  to  be  calculated  in  the 
same  manner  as  in  a  beam,  already  explained. 

Having  located  the  loads  and  their  reactions  upon  the 
vertical  line  af,  obtain  the  stresses  in  the  members  around 
the  joint  A  B  G  Z  from  the  point  b,  by  drawing  a  line  bg, 
parallel  to  B  G  in  the  stress  diagram ;  then  from  z  draw  the 
line  gz  parallel  to  G  Z,  and  the  intersection  of  these  two 
lines  will  be  the  point  g.  The  polygon  of  forces  around  this 
joint  is  from  b  to  g,  from  g  to  xr,  from  z  to  <7,  and  then  from 
tfback  to  b,  the  starting  point.  Bear  in  mind  that  the  forces 
in  the  stress  diagram  representing  the  reactions  must  have 
the  same  direction  as  the  reactions  in  the  frame  diagram. 
The  lines  determining  the  stresses  around  the  joint  R  C H  G 
should  next  be  drawn ;  b  c  having  been  determined,  from  c 
draw  a  line  c  h  parallel  to  C  H,  and  from  g  draw  a  line  gh 
parallel  toHG,  the  intersection  of  these  two  lines  determin- 
ing the  point  // ;  trie  polygon  of  forces  is  from  b  to  r,  from 
c  to  /i,  from  //  to  g,  and  back  again  from  g  to  b. 

Now  work  around  the  joint  C  D I  H;  cd  being  already 
known,  from  the  point  d  draw  the  line  d  i  parallel  to  D  /, 
and  from  the  point  h  draw  the  line  //  i  parallel  to  ///,  the 
intersection  of  the  two  lines  being  the  point  i.  In  going, 
around  the  joint  G  H IJ  Z,  the  stresses  in  the  members  sg, 
gh,  hi  have  already  been  determined  and  drawn  in  the 
stress  diagram.  Then  from  the  point  i  draw  the  line  ij 
parallel  to  //,  and  from  z  draw  the  line  zj  parallel  to  J Z, 
and  the  intersection  of  these  two  lines  will  be  the  point/. 
The  polygon  of  forces  around  this  joint  is  from  z  to  g,  from 
g  to  h,  from  h  to  z,  from  i  to  /,  and  from/  back  to  s,  the 
starting  point. 

The  next  joint  to  analyze,  in  going  around  the  truss,  is 
DEJ  I\ji,  id,  and  de  being  known,  the  only  remaining 
force  to  determine  is  the  stress  in  the  member  EJ.  The 


§  5  ARCHITECTURAL  ENGINEERING.  129 

point  c  being-  fixed,  draw  the  line  cj  parallel  to  RJ  in  the 
frame  diagram,  and  if  this  line,  which  completes  the  dia- 
gram, passes  through  the  pointy',  the  diagram  is  correct  and 
accurately  drawn.  The  stresses  around  the  right-hand  heel 
of  the  truss  are  all  known,  the  line  cj  just  drawn  having 
been  the  only  unknown  member  at  this  joint. 

The  polygon  of  forces  around  the  joint  D  RJ  I  is  from  d 
to  e,  from  c  to/,  from/ to  /,  and  from  /  to  d,  the  starting 
point.  The  polygon  of  forces  around  the  joint  R  F ZJ  is 
from  c  to  /",  from  f  to  ,?,  from  z  to/,  and  from/  back  again 
to  c,  completing  the  stress  diagram  for  the  dead  or  vertical 
load  upon  the  roof  truss. 

13O.  The  wind  diagram,  however,  remains  to  be 
drawn.  The  student  may  redraw  the  frame  diagram  as 
shown  in  Fig.  93.  The  wind  is  always  considered  as  acting 
normally,  or  at  right  angles,  to  the  roof,  the  amount  of  its 
pressure  at  the  different  joints  of  the  truss  being  shown  on 
the  frame  diagram.  As  the  heels  of  the  trusses  are  fixed, 
the  reactions  act  in  lines  parallel  to  the  wind  pressure. 

To  estimate  the   magnitude   of  the   reactions  /v,   and  A'.,, 
consider  the  left-hand   rafter   member  as   a  beam,  and  A, 
and  A.2  as  the  reactions  supporting  it.      The  moments  due  to 
the  wind  pressure  B  C  and  C  D  acting  about  Rl  are : 
at  B  C,  0,550  X  19.5     =  1  2  7,8  4  2  ft.-lb. 
at  CD,  1,950  X  29.00  =      5  7,8  3  7  ft.-lb. 
Total, 1  8  5,0  7  9  ft.-lb. 

The  lever  arm  with  which  A0  resists  the  wind  pressure 
acting  at  the  joints  is  29  feet  8  inches,  so  185,079 -=- 29.00 
=  0,200  pounds,  the  amount  of  the  reaction  at  A'.,.  The 
sum  of  the  loads  being-  4,000  +  0,550  +  1,950  =  13,100 
pounds,  the  reaction  at  A3,  is  13,100  —  0,200  =  0,840  pounds. 

First,  draw  the  load  line  a  d  in  the  stress  diagram,  Fig. 
94;  this  line  is  parallel  to  the  direction  of  the  wind  pressure 
and  the  reactions.  Now  lay  off  to  the  scale  to  which  the 
stress  diagram  is  drawn — in  this  case  2,000  pounds  to  every 
\  inch — the  force  a  b  equal  to  A  B  in  the  frame  diagram ; 
then  lay  off  be  equal  to  B  C  and  c  d  equal  to  CD.  From 


130 


ARCHITECTURAL  ENGINEERING. 


d  lay  off  the  magnitude  of  the  reaction  R^  or  d  z,  which 
determines  the  point  s,  and  the  distance  sa,  according  to 
scale,  represents  the  left-hand  reaction  R^  The  polygon  of 
external  forces  is,  then,  from  a  to  #,  from  b  to  c,  from  c  to  d, 
from  d  to  z,  and  from  z  back  again  to  a,  the  starting  point. 
This  polygon,  as  may  be  readily  seen,  is  a  straight  line,  as 
in  all  cases  so  far  analyzed. 

Continue  the  stress  diagram  Fig.  94,  by  going  around  the 
joint  A  B  G  Z\  from  the  point  b  draw  the  line  bg,  and  from 


FIG.  93. 


the  point  z  draw  the  line  sg  parallel  to  the  corresponding 
members  in  the  frame  diagram,  the  point  where  the  two 
lines  intersect  being  g.  The  polygon  of  forces  around  this 
point  is  from  a  to  b,  from  b  to  g,  from  g  to  z,  and  from  z 
back  again  to  a. 

The  next  joint  is  B  C H G;  be  has  been  already  obtained; 
then  from  the  point  c  draw  the  line  ch  parallel  to  C  H,  and 
from^draw  the  lmeg/i  parallel  to  H  G,  h  being  the  point 
where  these  two  lines  cross.  Disregard  the  two  members  in 
the  frame  diagram  shown  in  dotted  lines,  which  do  nothing 
towards  sustaining  the  wind  pressure.  Now  work  around 


ARCHITECTURAL  ENGINEERING. 


131 


the  joint  H  C D Z;  c  d  and  dz  are  known;  draw  from  z  the 
line  z  ky  parallel  with  Z  H  in  the  frame  diagram;  if  this 
closing  line  of  the  diagram  passes  through  the  point  //,  the 
diagram  has  been  drawn  accurately.  The  polygon  of  forces 
around  the  joint  B  C  H  G  is  from  b  to  c,  from  c  to  //,  from 
h  to  g,  and  g  back  again  to  b.  The  polygon  of  forces  around 
the  joint  C  D  Z  H  is  from  c  to  c/,  from  d  to  z,  from  z  to  h, 


8caU:  2000 


and  from  h  back  again  to  c.  The  polygon  of  forces  around 
the  joint  Z  G  H  is  from  z  to  £•,  from  ^  to  //,  and  from  h  to  z, 
the  starting  point. 

The  stress  diagram  for  both  the  dead  and  the  wind  load 
being  complete,  to  obtain  the  stress  in  each  member  of  the 
truss,  it  is  required  to  determine,  by  scale,  the  stress  due  to 
both  the  dead  and  wind  loads  in  each  member,  adding  the 
two  together  for  the  maximum  load  in  the  member.  To 
determine,  for  instance,  the  stress  in  the  strut  //  Gt  measure 

1-21 


132  ARCHITECTURAL  ENGINEERING. 


the  length  of  the  line  hg  in  the  stress  diagram,  Fig.  92,  for 

Si 
1- '-I/OS 


'9i  WAI 


the  dead  load;  then  measure  the  same  line  hg  in  the  stress 
diagram  for  the  wind,  Fig.  94,  add  the  two  measurements 


ARCHITECTURAL  ENGINEERING. 


133 


together,  and  determine  the  maximum  stress  in  the  strut  Jig 
from  the  assumed  scale  of  the  drawing. 

It  must  be  remembered  that  while  the  wind  acting  on  one 
side  of  a  truss  does  not  create  stresses  in  all  the  members  on 
the  opposite  side,  these  members  should  be  proportioned  in 
like  manner  as  the  other  members,  because  the  wind  is  quite 
as  likely  to  blow  upon  this  side  of  the  roof  and  reverse  the 
conditions. 

131.  Wooden  Truss  With  an  80-Foot  Span.— Fig.  95 
is  the  frame  diagram  for  an  80-foot  span  wood  roof  truss. 


Scale:  4OOO  Ib.to  a  inch 


FIG.  96. 


It  is  desired  to  draw  the  dead-load  diagram  and  the  wind- 
stress  diagram ;  also  to  design  and  properly  proportion  the 
roof  truss  to  resist  the  stresses  that  the  various  members 
may  be  required  to  sustain. 


134  ARCHITECTURAL  ENGINEERING.  §5 

Draw  the  frame  diagram  shown  in  Fig.  95,  and  mark  the 


dead  load  coming  upon  the  different  panel  points, -or  joints, 
in  the  truss.     The  truss  being  symmetrically  loaded,   the 


§5  ARCHITECTURAL  ENGINEERING.  135 

reactions  R1  and  AJa  are  each  equal  to  half  the  load  upon  the 
truss. 

Draw  the  stress  diagram,  Fig.  96,  for  the  dead  load,  say  to 
the  scale  of  4,000  pounds  to  -^  inch.  Draw  the  vertical  load 
line  aj\  and  determine  the  point  z,  having  previously  located 
upon  the  line  all  the  loads.  Then  draw  the  stress  diagram 
by  the  methods  previously  given.  Only  one-half  of  the 
diagram  need  be  drawn,  as  the  stresses  obtained  on  one  side 
of  the  center  of  the  truss  apply  to  the  other  side.  For 
instance,  fq  is  the  same  as  p  c.  Having  completed  half  of 
the  stress  diagram  for  the  dead  load,  the  student  should 
redraw  the  frame  diagram  as  shown  in  Fig.  97.  The  direc- 
tion and  amount  of  the  wind  pressure  at  the  several  panel 
points,  or  joints,  of  the  truss,  are  shown  in  the  frame  diagram. 

As  both  ends  of  the  truss  are  secured  against  sliding,  the 
reactions  act  in  a  direction  parallel  to  the  wind  pressure.  If 
the  left-hand  side  of  the  truss  be  secured,  the  right-hand  side 
being  on  rollers,  as  is  sometimes  the  case  with  iron  or  struc- 
tural steel  trusses,  to  allow  for  expansion,  then  the  right-hand 
reaction,  instead  of  being  parallel  to  the  direction  of  the  wind, 
would  be  vertical.  This  makes  considerable  difference  in  the 
stress  diagram,  as  will  be  explained  further  on. 

To  determine  the  magnitude  of  the  reactions  R1  and  Ry, 
let  A*,,  Fig.  97,  be  the  center  around  which  the  moment  of 
Ry  is  taken;  then  the  perpendicular  distance  between  the 
line  of  action  of  R^  and  the  point  Rl  will  be  71.22  feet. 
Extend  the  left-hand  rafter  until  it  cuts  the  line  of  action  of 
the  force  R^  at  the  point  jj/'.  Regard  this  extension  and  the 
rafter  as  a  beam,  and  calculate  the  magnitude  of  the  reactions 
Rl  and  R^  by  the  methods  given  for  beams. 

The  moments  about  Rl  are  as  follows : 

2,800X11.18  =  3  1,3  04  ft.-lb. 

2,800X22.36  =  6  2,6  0  8  ft.-lb. 

2,800X33.54  =  9  3,9  1  2  ft.-lb. 

1,400x44.72  =  6  2,6  0  8  ft.-lb. 

Total,    .   .  2  5  0,4  3  2  ft.-lb., 

and  250, 432-H7 1.22  =  3,516  pounds,  the  reaction  A'.    Having 


136 


ARCHITECTURAL  ENGINEERING. 


found  Aa,  find  A',  by  subtracting  A\  from  the  sum  of  the 
loads. 

The  sum  of  the  loads  is  1,400  +  2,800  +  2,800  +  2,800 
+  1,400  =  11,200  pounds.  Then  11,200-3,516  =  7,684 
pounds,  the  reaction  A*,. 

Next,  lay  out  the  wind  diagram,  Fig.  98,  by  drawing  the 
load  line  af  parallel  to  the  direction  of  the  wind  in  the  frame 
diagram,  Fig.  97.  Lay  off  to  the  scale  (in  this  case  4,000 
pounds  to  1  inch)  the  forces  ab,  be,  c  d,  de,  and  ef  equal  to 
A  B,  B  C,  CD,  and  so  on,  in  the  frame  diagram.  Then 


Scale  4OOO  Ib.  to  1  inch. 

P 

FIG.  98. 

from  a  lay  off  az  equal  to  the  reaction  ZA,  or  Rt.  If 
the  other  forces  or  loads  have  been  laid  off  accurately, 
fz  should,  upon  measurement,  be  found  equal  to  the  right- 
hand  reaction  R^. 

The  first  joint  to  analyze  is  A  B  K  Z,  Start  at  b  and  draw 
the  line  b  k  parallel  to  B  K  in  the  frame  diagram ;  then  from 
z  draw  z  k  parallel  to  K  Z,  where  the  two  lines  intersect  being 
the  point  k.  Then  the  polygon  of  forces  around  this  joint 


§  5  ARCHITECTURAL  ENGINEERING.  137 

is  from  a  to  /;,  from  b  to  /•,  from  k  to  z,  and  from  z  back 
again  to  the  starting  point  a. 

The  next  joint  to  analyze  is  B  C L  K.  From  the  point  c 
draw  the  line  c  I  parallel  to  C  L  in  the  frame  diagram,  and 
from  k  draw  the  line  /'  /  parallel  to  the  member  L  K,  the 
point  of  intersection  being  /.  The  polygon  of  forces  around 
this  joint  is  from  b  to  c,  from  c  to  /,  from  /  to  /-,  and  from  k 
back  again  to  b. 

To  analyze  the  joint  KLMZ:  kl  being  already  known, 
the  next  number  is  L  J/;  therefore,  from  the  point  /  draw 
the  line  /;//  parallel  to  L  J/in  the  frame  diagram.  As  the 
next  member  around  this  joint  is  JfZ,  to  which  j/iz  in  the 
stress  diagram  is  parallel,  the  point  m  is  located  where  the 
line  /;//  intersects  the  line  niz;  this  completes  this  joint, 
the  polygon  of  forces  around  it  being  from  k  to  /,  from 
/  to  m,  from  111  to  s,  and  from  z  back  again  to  /'. 

To  determine  the  stresses  in  the  members  around  the 
joint  C D  N ML,  draw  from  the  point  d  the  line  dn,  parallel 
to  the  member  DN  in  the  frame  diagram;  then,  from 
the  point  m  draw  ;;/  n  parallel  to  N  M.  The  polygon  of 
forces  around  this  joint  is  from  c  to  d,  from  d  to  n,  from  n 
to  »/,  from  ;//  to  /,  and  from  /  back  again  to  c . 

To  analyze  the  forces  around  the  joint  M  N  O  Z,  draw  from 
n  the  line  n  o  upwards,  parallel  to  N  O  in  the  frame  diagram  ; 
as  the  next  member  O  Z  is  horizontal,  the  point  o  must  be 
at  the  intersection  ;/  o  and  o  z.  This  completes  this  joint,  and 
the  polygon  of  forces  around  it  is  from  ;//  to  o,  from  n  to  o, 
from  o  to  z,  and  from  z  back  again  to  ;//,  the  starting  point. 

Now  analyze  the  joint  D E PO N.  From  c  draw  the  line 
ep  parallel  to  E  P  in  the  frame  diagram,  and  from  o  draw 
the  line  op  parallel  to  P  O.  The  intersection  of  these  two 
lines  determines  the  point  /,  and  the  polygon  of  forces 
around  this  joint  is  from  d  to  c\  from  c  to  /,  from  /  to  o, 
from  o  to  «,  and  from  ;/  back  again  to  </,  the  starting  point. 

The  analysis  of  the  joint  O  P  Q  Z  is  made  by  drawing  the 
vertical  line / q  from  the  point/;  the  point  where pq  inter- 
sects q  z  is  q.  Then  the  polygon  of  forces  around  this  joint 
is  from  o  to  p,  from  /  to  g,  from  q  to  z,  and  from  z  back 


138 


ARCHITECTURAL  ENGINEERING. 


again  to  o.  The  members  shown  in  dotted  lines  do  not  sus- 
tain any  stresses  from  the  pressure  of  the  wind,  when  it 
blows  upon  the  left-hand  side  of  the  truss. 

The  final  joint  to  consider,  thus  completing  the  stress 
diagram,  is  E  F  Q  P.  There  is  only  one  unknown  force 
around  this  joint,  and  that  is  the  stress  in  the  member  F  Q. 
A  line  drawn  from /in  the  stress  diagram,  parallel  with  the 
member  F  Q,  should  pass  through  the  point  q;  if  it  does 
not,  the  diagram  has  been  inaccurately  drawn.  This  is 
always  a  test  of  the  accuracy  of  the  stress  diagram,  and  if  the 
last  line  in  this  diagram  does  not  close  on  the  proper  point, 
when  drawn  parallel  to  the  member  it  represents,  there  is 
something  so  radically  wrong  as  to  demand  that  the  stress 
diagram  be  redrawn,  to  determine  whether  the  loads  and 
reactions  have  been  laid  out  correctly,  and  whether  any  of  the 
joints  or  members  in  the  structure  have  been  passed  over. 

132.  The  two  diagrams  completed,  measure  the  differ- 
ent lines  and  obtain  the  stresses  in  the  various  members, 
tabulating  the  results,  as  follows : 


Member. 

Dead-Load 
Diagram.  - 

Wind-Load 
Diagram. 

Total  of  Both. 

Kind  of  Stress. 

BK 

27,000 

12,000 

39,000 

Compressive. 

CL 

23,500 

10,000 

33,500 

Compressive. 

DN 

19,500 

7,600 

27,100 

Compressive. 

EP 

16,000 

5,680 

21,680 

Compressive. 

KZ 

24,000 

13,300 

37,300 

Tensile. 

MZ 

21,000 

10,000 

31,000 

Tensile. 

OZ 

17,500 

7,000 

24,500 

Tensile. 

•LK 

4,000 

3,500 

7,500 

Compressive. 

N  M 

5,000 

4,400 

9,400 

Compressive. 

PO 

6,500 

5,400 

11,9.00 

Compressive. 

ML 

1,600 

1,500 

3,100 

Tensile. 

ON 

3,500 

3,000 

6,500 

Tensile. 

QP 

10,600 

4,500 

15,100 

Tensile. 

FQ 

16,000 

6,600 

22,600 

Compressive. 

The  values  in  the  above  table  represent  the  stress  in 
round  numbers  upon  the  various  members  in  the  truss,  as 
obtained  from  the  dead-load  and  wind  diagrams. 


§  5  ARCHITECTURAL  ENGINEERING.  139 

The  size  of  the  timber  and  tension  rods  in  the  truss  may 
now  be  calculated. 

133.     Designing   the  Members  of  tlie  Truss. — The 

first  vertical  tension  member  is  J//,,  this  member  having  a 
total  pull  upon  it  of  3,100  pounds.  This  rod  will  be  made 
of  wrought  iron,  the  tensile  strength  of  which  is  52,000 
pounds.  If  a  factor  of  safety  of  4  is  used,  the  safe  load  will 
be  52, 000  -=-4  =  13,000  pounds  per  square  inch.  A  rod  £ 
inch  in  diameter  has  an  area  at  the  root  of  the  screw  thread 
of  .  302  square  inch.  Then  the  safe  strength  of  a  rod  f  inch 
in  diameter,  threaded  at  the  ends,  is  .302  X  13,000  =  3,920 
pounds.  This  member  being  required  to  support  3,100 
pounds  only,  a  wrought-iron  rod  f  inch  in  diameter  is  amply 
strong.  The  tension  member  O  N  is  subjected  to  a  stress 
of  6,500  pounds.  A  rod  1  inch  in  diameter  has  an  area  at 
the  root  of  the  thread  of  .550  square  inch.  Hence,  its  safe 
strength  is  .550x13,000  —  7,150  pounds,  or  slightly  in 
excess  of  that  required.  The  member  Q  P  is  subjected  to  a 
stress  of  15,100  pounds.  The  area  at  the  root  of  the  thread 
of  a  rod  1|  inches  in  diameter  is  1.29  square  inches,  and  the 
safe  strength  is  1.29x13,000  =  16,770  pounds,  which 
exceeds  the  strength  required. 

To  calculate  the  size  of  timber  demanded  for  the  rafter 
member,  note  that  it  is  usual  in  timber  trusses  to  make  the 
rafter  members  of  one  size  and  of  one  length.  This  truss 
requires,  then,  a  large  stick  of  timber,  say  about  45  feet 
long,  for  the  rafter  members,  and  obtainable,  likely,  by 
special  order.  As  the  maximum  stress  on  the  rafter,  made 
of  one  piece  of  timber  throughout,  is  at  B  K,  we  have  to 
calculate  its  size  at  this  point  only.  Assume  that  the  truss 
supports  heavy  purlins  only  at  the  panel  points,  or  joints, 
of  the  truss,  common  rafters  being  laid  up  and  down  the 
roof,  resting  on  these  purlins  as  shown  in  Fig.  99.  This 
concentrates  all  the  load  on  the  truss  at  the  panel  points, 
and  B  K  is  consequently  not  to  be  estimated  as  a  beam, 
sustaining  as  it  does  only  the  compressive  stress  as  obtained 
by  the  diagram. 


140  ARCHITECTURAL  ENGINEERING.  §  5 

The  total  compressive  stress  in  B  K  is  39,000  pounds,  and 
its  length  is  11.18  feet  =  134  inches,  nearly.  Using  an 
8"x8"  yellow-pine  timber,  the  ultimate  compressive 
strength,  parallel  to  the  grain,  as  given  in  Table  6,  Art.  61, 
is  4,400  pounds  per  square  inch.  By  formula  3,  Art.  7O, 
the  ultimate  compressive  breaking  strength  of  B  K  as  a 

column  is 

4,400X134 


100X8 
=  4,400  —  737  =  3,663  pounds  per  square  inch. 

If,  on  account  of  its  uncertain  nature,  a  factor  of  safety  of  5 
is  used  for  the  timber,  the  safe  bearing  value  of  the  column 
is,  then,  3,663-^5  =  733  pounds  per  square  inch.  The  area 
of  an  8"  X  8"  column  is  64  square  inches,  which  multiplied 
by  733  gives  a  safe  load  of  46,912  pounds.  While  consider- 
ably in  excess  of  the  39,000  pounds  required,  this  is  the  nearest 
even-sized  timber  that  could  be  used  for  this  member. 

To  determine  the  size  of  timber  required  for  the  tie- 
member,  bear  in  mind  that  the  greatest  pull,  or  tensile 
stress,  upon  this  member  is  at  KZ,  and  amounts  to  37,300 
pounds.  This  member,  made  of  yellow-pine  timber,  must 
be  spliced,  say  at  the  center.  It  is  safe  to  assume  that,  in 
making  the  connection  and  joints,  one-third  of  this  timber 
will  be  cut  away.  The  ultimate  tensile  strength  of  yellow 
pine  is,  according  to  Table  6,  Art.  61,  8,000  pounds  per 
square  inch.  This  divided  by  the  factor  of  safety  of  5  equals 
1,600  pounds,  the  safe  stress  that  a  square  inch  will  sustain. 

For  convenience  in  construction,  a  6"x8"  timber  is 
selected.  The  area  of  this  timber  is  48  square  inches  ;  f  of 
this  is  32  square  inches.  Then,  32  X  1,600  =  51,200  pounds. 
A  6"  X  8"  timber  is  thus  amply  strong,  but  upon  drawing 
the  truss  and  laying  out  the  detail  of  its  connections,  it  may 
be  found  necessary  to  use  a  timber  8  in.  x  8  in.  In  deter- 
mining the  size  of  the  member  P  O,  bear  in  mind  that  the 
stress  on  this  member  is  11,900  pounds.  The  member  P  O 
in  the  frame  diagram,  Fig.  97,  measures  about  18  feet. 
The  formula  for  wood  columns  shows  that  this  member 
may  be  made  of  a  6"x8"  timber,  which  is  amply  strong. 


- 

—  ^*— 
•a.           jc 

i 

- 

/^* 
a.           a 

•n. 

£ 

/** 
a           ji 

1 

^ 

/fc* 
a.           i 

k 

4 

V 

~^ 

X1 

! 

17 

•*s 

<" 

-s- 

i 

i 

1 

— 

\ 

,-' 

i 

Shea  thin 


Detail  of  Splice  in  Tie  Member. 

Purlins.^ 


8   did. MOd 

Wrt .  Iron. 


^8X8" Yellow  fine. 


ft — 


Screws. 

Detail  of  J( 

FIG. 


Wrt.Iron  Washer  1  Thick 


ix8  Yellow  Pine. 


-4X#"Yellow  Pine. 


la.  JRod 
't.Iron. 


1  diet.  Mod 
Wrt.  Iron. 


Wrt.Iron  Washer  8X8? 


^.~ 
Splice  Plate  not  shown'. 

fs  on  Tie  Member. 


§  5  ARCHITECTURAL  ENGINEERING.  1-11 

A  4"x8"  timber  would  probably  carry  the  load,  but  the 
length  of  the  strut  being  more  than  45  times  the  width  of 
the  least  side,  the  next  larger  stock  size  of  timber  is  adopted, 
viz.,  G  in.  X  8  in.  The  timbers,  to  facilitate  making  the  con- 
nections, should  all  be  of  one  thickness. 

In  this  case,  for  instance,  the  face  of  all  the  members  in 
the  truss  are  flush,  for  all  are  of  the  same  thickness.  The 
other  struts,  K L  and  J/Ar,  will  be  found,  in  like  manner,  to 
be  amply  strong,  if  made  of  -4"x8"  timber.  The  tic-member 
is  of  such  a  length  as  to  be  made  up  of  at  least  two  pieces, 
spliced  at  the  center.  The  shear  of  the  wood  parallel  to  the 
grain  determines  the  strength  of  the  splice,  as  explained  in 
the  example,  Art.  62.  The  bolts  may  be  used  to  hold  the 
splice  together,  but  should  not  be  depended  upon  to  with- 
stand any  of  the  pull  on  the  tie-member. 

In  estimating  the  strength  of  the  tie-member  at  the  heel, 
it  may  be  difficult  to  obtain  a  sufficient  section  of  wood  at 
the  end  of  the  tie  to  resist  the  shear  parallel  to  the  grain  on 
the  line  a  b.  In  this  instance,  the  pull  on  the  member  is 
37,300  pounds.  The  area  of  the  shearing  section  on  the  line 
a  b  is  8  X  26  =  208  square  inches.  According  to  Table  G, 
Art.  61,  the  ultimate  shearing  strength  of  yellow  pine  per 
square  inch,  parallel  to  the  grain,  is  400  pounds.  If  a  factor 
of  safety  as  low  as  4  is  used,  the  safe  strength  per  square 
inch  will  be  100  pounds.  Then,  208  X  100  =  20,800  pounds. 
Now,  as  previously  stated,  the  stress  is  37,300  pounds,  and 
the  remaining  stress  of  1C,  500  pounds  will  have  to  be  taken 
up  by  securely  bolting  the  joint  and  by  strapping  it,  as  shown 
in  Fig.  99.  It  is  seldom  found  that  sufficient  shear  can  be 
obtained  in  the  wood  to  resist  the  stress  at  this  point,  it 
being  always  necessary  to  depend  more  or  less  on  the  bolts. 
In  a  large  truss,  like  that  under  consideration,  the  best  prac- 
tice demands  that  the  wrought-iron  strap  be  so  proportioned 
as  to  be  capable  of  resisting  all  the  stress  not  borne  by  the 
shear  of  the.  wood  on  the  line  a  b,  no  reliance  whatever 
being  placed  on  the  bolts.  The  lagscrews  simply  retain  the 
straps  in  position.  One  notch  in  the  tie,  to  receive  the 
rafter,  is  always  preferable  to  two,  as  given  in  Fig.  100. 


142  ARCHITECTURAL  ENGINEERING.  §  5 

Nothing  is  gained  in  strength  by  using  two,  because  one  or 
the  other  of  the  joints  will  open  from  shrinkage,  and,  conse- 
quently, either  the  piece  h  or  i  will  shear  off  before  any 
stress  is  brought  upon  the  other. 


FIG.  100. 

Care  should  be  taken  that  the  washers  are  of  sufficient 
area  to  prevent  them  from  cutting  into  the  wood.  Table  6, 
giving  the  allowable  compression  perpendicular  to  the  grain, 
is  of  use  in  determining  the  sizes  of  the  washers. 

The  tension  member  f  inch  in  diameter,  Fig.  99,  is  not 
needed  to  resist  any  stress,  but  is  useful  in  the  truss  to  hold 
the  strut  in  position  and  prevent  trouble  from  any  shrinkage 
that  may  occur.  No  further  details  need  be  given  in  the 
design  of  wood  roof  trusses,  good  judgment  and  foresight  on 
the  part  of  the  designer  securing,  with  the  principles  and 
practice  already  laid  down,  the  desired  result,  viz.,  maximum 
strength  in  the  structure,  with  a  minimum  amount  of 
material. 


DESIGN  FOB  A  LARGE  BUILDING. 

134.  The  combination  of  general  principles  and  practi- 
cal rules  and  formulas  heretofore  presented,  is  sufficient  to 
enable  the  student  to  solve  any  problem  involved  in  the 
design  of  a  large  building  having  floors  supported  by 
masonry  walls  and  wooden  or  cast-iron  columns  and  a  roof 


§  5  ARCHITECTURAL  ENGINEERING.  143 

carried  by  timber  trusses.  A  practical  example  of  the  appli- 
cation of  the  preceding  rules  and  formulas  is  now  given  in 
the  form  of  a  complete  design  for  a  five-story  factory  build- 
ing, with  basement. 

The  foundations  are  of  rough  stone  in  cement  mortar. 
The  soil  on  which  the  building  is  to  rest  is  of  dry  clay.  The 
windows  are  to  be  4  feet  wide  and  the  floor  girders  8  feet 
from  center  to  center,  while  the  basement  columns  are  to  be 
of  cast  iron,  other  columns  and  posts  being  of  yellow  pine. 
The  first-floor  girders  are  of  steel,  with  steel  floorbeams 
placed  on  4-foot  centers.  A  4-inch  brick  arch  is  built 
between  the  beams  and  is  filled  in  with  concrete,  having  a 
cement  top  finishing  coat.  The  other  floors  in  the  building 
are  supported  upon  yellow-pine  trussed  beams,  the  flooring 
to  be  of  3-inch  yellow-pine  plank,  splined  and  grooved  with 
1-inch  finished  flooring  upon  the  top. 

The  live  load  upon  the  floor  will  be  120  pounds.  The  roof 
will  be  supported  upon  a  wood  truss  making  a  clear  span  of 
the  top  floor,  and  the  roof  covering  is  slag  upon  2-inch 
yellow-pine  planks.  The  pitch  of  the  roof  is  4  inches  per 
foot.  The  brick  walls  of  the  building  are  laid  in  lime  mortar. 

Fig.  101  is  a  scale  drawing  of  a  section  and  part  elevation 
of  the  building.  Fig.  102,  showing  details  of  the  various 
members  and  connections,  is,  in  fact,  a  complete  working 
drawing  of  the  entire  building.  The  4-inch  brick  arch 
between  the  steel  beams  supports  all  the  load  a  floor  of  this 
character  may  ever  require  to  carry.  The  bluestone  caps 
upon  which  the  columns  rest  are  17  inches  thick,  or  equal  in 
thickness  to  one-half  the  width  of  the  side.  This  is  accord- 
ing to  the  rule  previously  given  in  designing  foundation 
piers  for  columns.  It  is,  however,  usual  to  find  capstones 
made  much  thinner  in  ordinary  work,  but,  if  good  results 
are  desired,  adhere  to  this  rule.  The  cast-iron  caps  and 
wall  boxes  shown  in  detail  in  Fig.  103  are  known  as  the 
Goetz-Mitchell  wall  box  and  column  cap. 

The  wall  box  is  especially  good  as  an  anchor  to  bind  the 
wall  and  the  end  of  the  girder  together.  The  cast-iron 
dowel-pins  on  the  casting,  over  which  the  wood  girder  is 


144  ARCHITECTURAL  ENGINEERING. 


5 


placed,  securely  hold  the  girder  to  the  box ;  the  flared  sides 
of  the  box,  when  built  into  the  wall,  prevent  it  from  being 


Side  Elevation. 


Foundation  Footings  -  Cement  Mortar. 


Sectional  Elevation. 
FIG.  101. 


pulled  out.  Vent  holes  along  the  edge  of  the  wall  box 
ventilate  the  end*  of  the  wood  girder  and  thus  prevent  any 
tendency  to  dry  rot. 


/' ' Harct '  Map/e  Floorimr  Crossed 
P/ne 


DETAILS  OF  WOOD  G/RDERS 

OTHER  FLOORS  SIMILAR 


Concrete  ^  Peep. 

DETAILS  Of  BU/LD 


ffoucrfc  Sfcne  Foundation, 


'         ffouyf*  S?om  F 


ARCHITECTURAL  ENGINEERING. 


145 


The  details  shown  in  Fig.  102  are  good  practical  examples 
of  the  application  of  the  principles  already  set  forth ;  the 
student  is  urged  to  study  them  carefully  and  compare  their 
dimensions  with  the  values  obtained  by  the  use  of  the  rules. 

t~  Detail  for  trail  Boxes. 


2nd  Floor.  3rd  &  4th  Floor.  5th  Floor. 
Details  for  Column    Caps. 

FIG.  103. 

He  must  constantly  bear  in  mind  that  the  engineer  often 
meets  conditions  to  which  no  published  rule  or  formula 
applies;  these  conditions  can  be  successfully  met  only  by  the 
exercise  of  good  judgment  in  applying  the  general  principles 
of  mechanics,  coupled  with  a  thorough  and  practical  knowl- 
edge of  the  properties  and  uses  of  available  structural 
materials. 


ARCHITECTURAL  ENGINEERING. 

(CONTINUED.) 


PROPERTIES  OF  SECTIOXS. 


INTRODUCTORY. 

1.  The  calculations  involved  in  the  design  of  such 
"built-up"  members  of  a  building'  as  steel  columns  and 
plate  girders — members  that  are  formed  by  the  combination 
of  several  of  the  simple  sections  produced  by  the  rolling 
mills — require  a  knowledge  of  certain  mathematical  proper- 
ties of  the  simpler  sections,  together  with  the  methods  by 
which  these  properties  may  be  calculated.  In  many  cases 
the  exact  determination  of  the  required  properties  is  based 
on  complicated  mathematical  principles;  there  are,  how- 
ever, numerous  formulas  and  practical  methods  by  means 
of  which  the  values  for  all  sections  used  in  ordinary  practice 
may  be  determined,  either  exactly  or  with  a  degree  of 
approximation  sufficiently  close  for  all  practical  purposes. 


THE    XEUTRATj    AXIS. 

2.  location  of  the  Neutral  Axis. — In  Art.  92,  Archi- 
tectural Engineering,  §  5,  it  was  stated  that  the  neutral 
axis — the  line  separating  the  fibers  in  tension  from  those 
subjected  to  compression  in  a  section  of  a  beam — always 
passes  through  the  center  of  gravity  of  the  section.  It  is 
also  evident  from  what  was  there  stated  that  the  neutral 
axis  is  perpendicular  to  the  direction  in  which  the  load  acts 
on  the  beam ;  therefore,  to  find  the  neutral  axis  of  a  section 

1-22 


ARCHITECTURAL  ENGINEERING. 


with  reference  to  a  set  of  loads  applied  in  a  given  direction, 
it  is  only  necessary  to  pass  a  line  through  the  center  of 
gravity  perpendicular  to  the  direction  of  the  load. 

A  simple  approximate  method  of  locating-  the  center  of 
gravity  and  neutral  axis  of  a  section  is  shown  in  Fig.  1. 


FIG.  1. 

Draw  the  outline  of  the  section,  either  full  size,  or  to  some 
convenient  scale,  on  a  piece  of  heavy  cardboard.  Cut  the 
section  out  and  balance  it  carefully  over  a  knife  edge  as 
shown  in  the  figure;  the  line  along  which  it  rests  on  the 
edge  of  the  knife  is  a  line  passing  through  the  center  of 
gravity,  and  by  locating  two  such  lines  in  different  direc- 
tions, the  center  of  gravity  will  be  found  at  their  point  of 
intersection. 

3.  Locating  the  Neutral  Axis  toy  Means  of  the 
Principle  of  Moments. — A  convenient  method  of  locating 
the  neutral  axis  is  based  on  the  principle  that  the  moment 

of    any  figure,    with 
— 6  respect  to  a  given  line 
as  an  axis  or  origin  of 
*\+f_ i ?0  moments,  is  equal  to 

--w- 

ISL-I 


-LJ5 


~—  -».. i  -• 

-— /o" »| 

FIG.  2. 

in  Fig.  2,  take  the  line  a  b  as  the  line  of  origin  of  moments. 


the  sum  of  the  mo- 
ments of  its  separate 
parts  with  respect  to 
the  same  axis.  Thus, 


§  6  ARCHITECTURAL  ENGINEERING.  3 

Divide  the  figure  into  the  three  rectangles  ;;/,  ;;/',  in".  In 
accordance  with  the  principles  stated  in  Art.  #9,  ArcJii- 
tectural  Engineering,  g  5,  the  center  of  gravity  of  each  of 
these  rectangles  is  midway  between  its  edges;  the  distances 
of  the  respective  centers  from  the  axis  are,  therefore,  \\,  7, 
and  11^-  inches.  The  areas  of  the  figures  are  respectively 
3x3  =  !)  square  inches,  8x1=8  square  inches,  and  10 x  1 
=  10  square  inches.  The  moments  of  these  areas  about 
the  axis  a  b  are : 

Section  m      9x    \\  =      1  3.5 

Section  ;//'     8x7=      5  G.O 

Section  ;//"  10xll£  =  1  1  5.0 

Total,     .     .     .      184.5 

The  area  of  the  whole  section  is  equal  to  0  -(-  8  -(-  10  =  27 
square  inches,  the  sum  of  the  areas  of  the  rectangles;  the 
distance  c  of  its  neutral  axis  from  the  line  of  the  origin  of 
moments  is,  therefore,  184.5-4-27  =  G.83  inches,  or  nearly 
6|£  inches. 

O   £i 

It  is  not  necessary  that  the  line  of  the  origin  of  moments 
should  coincide  with  an  edge  of  the  figure  as  in  Fig.  2,  since 
any  other  line  parallel  with  the  direction  of  the  required 
neutral  axis  gives  the  same  results;  in  most  cases,  however, 
it  will  be  found  more  convenient  to  take  the  axis  about 
which  the  moments  are  calculated  on  one  of  the  extreme 
edges  of  the  section. 

Since  the  section  shown  in  Fig.  2  is  symmetrical  with 
respect  to  an  axis  perpendicular  to  the  neutral  axis,  it  is 
evident  that  its  center  of  gravity  is  on  their  intersection. 
If,  however,  there  were  no  axis  of  symmetry,  the  center  of 
gravity  could  have  been  located  by  taking  a  second  line 
perpendicular  to  a  b  as  an  origin  of  moments  and  finding 
the  neutral  axis  parallel  to  it.  The  intersection  of  this  neu- 
tral axis  with  the  one  first  found  is  the  center  of  gravity  of 
the  section.  In  accordance  with  the  principles  illustrated 
in  the  above  example,  we  have  the  following  rule : 

Rule. —  TofindtJic  neutral  axis  of  any  section,  first  divide  it 
into  a  number  of  simple  parts,  each  of  whose  areas  and  centers 


4  ARCHITECTURAL  ENGINEERING.  §  6 

of  gravity  can  be  readily  found ;  then  find  the  sum  of  the 
moments  of  t lie  areas  of  cacti  of  these  parts  with  respect  to  an 
axis  parallel  to  the  required  neutral  axis.  Finally,  divide 
this  sum  by  the  sum  of  the  areas  of  the  parts  of  the  section, 
and  the  result  will  be  the  perpendicular  distance  from  the 
axis  of  the  origin  of  moments  to  the  required  neutral  axis. 

4.     Application  of  the  Rule  to  a  Built-TJp  Section.— 

Fig.  3  shows  a  section  of  the  rafter  member  of  a  large  roof 
truss  formed  of  a  f'xlG"  web-plate  and  a  f"x!2"  flange 
plate,  the  two  joined  by  two  4"  X  4"  X  \"  angles.  What  is  the 
distance  of  the  neutral  axis  of  the  section  from  the  top  edge 
of  the  flange  plate  ?  By  means  of  the  principles  previously 


-ft 


FIG.  3. 


given,  the  centers  of  gravity  of  the  two  rectangular  plates 
are  easily  located  as  shown.  The  centers  of  gravity  of  the 
angles  might  also  be  located  by  applying  the  rule  given  in 
the  preceding  article;  this,  however,  is  unnecessary,  since 
the  center  of  gravity  can  be  obtained  directly  by  referring 
to  the  tables  of  the  properties  of  rolled  sections.  By 
referring  to  the  table,  "Properties  of  Angles  with  Equal 
Legs,"  the  center  of  gravity  of  a  4"x4"xf£"  angle  is 
found  to  be  1.29  inches  from  the  back  of  flange,  and  of  a 
4" X 4" X tV"  angle  the  distance  is  1.12  inches;  the  center  of 


§  6  ARCHITECTURAL  ENGINEERING.  5 

gravity  of  a  4"  x4"  xV'  angle  is,  therefore,  nearly  1^  inches 
from  the  back  of  a  flange,  thus  giving  us  the  distance 
H  +  f  =  If  inches  from  the  top  edge  of  the  flange  plate  to 
the  axis  through  the  centers  of  gravity  of  the  angles.  From 
the  table,  "Areas  of  Angles,"  it  is  also  found  that  the  area 
of  the  section  of  a  4"  X-i"  X  V  angle  is  3.75  square  inches. 

The  area  of  the  section  of  the  flange  plate  is  f  X  12  =  4.5 
square  inches,  and  of  the  web-plate,  fXlC  =  0  square 
inches;  the  area  of  the  whole  section  is,  therefore,  2x3.75 
4- 4. 5  +  6  =  18.0  square  inches. 

The  moments  of  the  areas  of  the  separate  sections,  with 
respect  to  the  line  a  b,  are  as  follows : 

Flange  plate        4. 5  X  •&  =        .84 

Two  angles  2  X  3. 75  X  If  =  1  2. 1  0 

Web-plate  6  X  8|  =  5  0.2  5 

Total, 6  3.2  8 

The  distance  c  from  the  top  edge  of  the  flange  plate  to  the 
neutral  axis  de  of  the  section  is,  therefore,  63. 28 -h  18.0 
=  3.51  inches. 

5.     Graphical  Method  of  Locating  the  Xetitral  Axis. 

Let  it  be  required  to  determine  the  position  of  the  neutral 
axis  of  the  cast-iron  beam  section  shown  in  Fig.  4.  First, 
divide  the  depth  of  the  section  into  any  number  of  parts — 
as  has  been  done  in  this  case  by  the  dotted  lines  w'  x,  y  z, 
etc. — whose  areas  and  centers  of  gravity  can  readily  be 
found.  Then  compute  the  area  and  locate  the  center. of 
gravity  of  each  part.  In  Fig.  4,  the  area  of  the  top  slice  is 
12  square  inches,  the  area  of  each  of  the  slices  in  the  web 
of  the  section  is  3  square  inches,  and  the  area  of  the  bottom 
flange  is  28  square  inches.  Assume  some  scale  whose  unit 
of  length  represents  1  square  inch  ;  for  example,  in  this  case, 
let  y^g-  of  an  inch  represent  1  square  inch.  Then,  commen- 
cing at  some  point  a,  lay  off  on  the  line  a  /,  lengths  a  b,  be, 
cf,  etc.  which  represent  the  respective  areas  of  the  succes- 
sive parts  into  which  the  section  of  the  beam  has  been  divided, 
beginning  at  the  top  part.  Thus,  with  the  scale  of  J-g-  =  1 


6  ARCHITECTURAL  ENGINEERING.  §  6 

square  inch,  the  line  a  6,  which  represents  an  area  of  12 
square  inches,  is  |f  =  f  inch  long;  each  of   the  lines  be, 


cf,  etc.  is  ^  inch  long ;  and  the  line  k  I  is  \ £  =  If  inches 
long. 


§  6  ARCHITECTURAL  ENGINEERING.  7 

From  the  points  a  and  /,  draw  45°  lines,  intersecting  at 
the  point  ;;/.  Then  from  the  points  b,  c,  f,  etc. ,  draw  the 
lines  b  in,  c ;//,  /;;/,  etc.  Throiigh  the  center  of  gravity  of 
each  of  the  parts  of  the  section,  draw  indefinite  lines  paral- 
lel to  a  I. 

From  the  point  ;/,  where  the  line  through  the  center  of 
gravity  of  the  top  section  intersects  the  line  a  m,  draw  the 
line  ;/  o  parallel  to  the  line  b  m  until  it  intersects  the  line 
passing  through  the  center  of  gravity  of  the  second  slice  in 
the  point  o ;  draw  the  line  op  parallel  to  c  in ;  p  q  parallel  to 
fm ;  q  r  parallel  to  g  in ;  r  s  parallel  to  // ;;/ ;  s  t  parallel  to 
im\  tu  parallel  toy";;/;  and  uv  parallel  to  kin. 

From  the  point  r,  which  is  the  point  at  the  intersection  of 
the  line  u  v  with  the  line  drawn  through  the  center  of  grav- 
ity of  the  last  elementary  section,  draw  the  line  v  iv  parallel 
to  tn  /;  then  its  intersection  u>  with  the  line  a  in  is  a  point  on 
the  required  neutral  axis.  If  the  line  a  in  is  so  short  that 
the  line  v  w  fails  to  cut  it,  it  may  be  extended  indefinitely, 
as  shown  at  mx'  so  as  to  make  it  intersect  with  the  line  v  w. 
Having  found  the  point  tc(,  draw  the  horizontal  line  dc 
through  it.  This  line  is  the  required  neutral  axis  of  the 
figure,  and  passes  through  its  center  of  gravity. 

This  method  of  determining  the  position  of  the  neutral 
axis  and  center  of  gravity  may  be  applied  to  any  irregular- 
shaped  section  whatever,  and  in  many  cases  may  be  found 
more  convenient  than  the  mathematical  method. 

When,  as  in  Fig.  4,  the  section  is  made  up  of  several  reg- 
ular parts  whose  centers  of  gravity  can  be  readily  located,  it 
is  not  necessary  to  subdivide  any  one  of  these  parts.  Thus, 
the  center  of  gravity  of  the  web  of  the  beam  is  located  on 
the  horizontal  line  through  r,  and  its  area  is  represented  by 
the  distance  b  k.  We  can  therefore  draw  from  n  a  line  par- 
allel to  b  m  until  it  intersects  the  horizontal  line  through  the 
center  of  gravity  of  the  web  member;  then,  from  this  point, 
draw  a  line  parallel  to  k  m  until  it  intersects  the  horizontal 
through  the  lower  section  at  the  point  v.  The  point  i\  as 
thus  located,  is  identical  with  the  point  previously  found 
when  the  web  section  was  divided  into  the  small  parts,  and 


8  ARCHITECTURAL  ENGINEERING.  §  6 

the  line  drawn  from  v  parallel  to  Im  until  it  intersects  a  m, 
locates  the  point  w  on  the  neutral  axis,  as  before.  If,  how- 
ever, the  center  of  gravity  of  the  web  cannot  be  readily 
located,  it  is  better  to  divide  it  into  small  parts,  as  in  the 
first  method. 


THE  MOMENT  OF  INERTIA. 

6.  The  term  moment  of  inertia  is  a  mathematical 
expression  which  depends  on  the  distribution  of  either  the 
material  of  a  body  or  the  area  of  a  surface  with  respect 
to  a  given  axis.  As  applied  to  the  area  of  a  plane  figure, 
the  moment  of  inertia,  with  respect  to  an  axis  lying  in  the 
same  plane,  is  numerically  equal  to  the  sum  of  the  products 
obtained  by  multiplying  each  of  the  elementary  areas  of 
which  the  figure  is  composed  by  the  square  of  its  distance 
from  the  given  axis. 

By  elementary  area  is  meant  an  area  smaller  than  any 
with  which  we  are  accustomed  to  deal  in  ordinary  calcula- 
tions ;  it  is.  therefore,  impossible  to  find  an  exact  expression 
for  the  moment  of  inertia  of  a  figure  by  the  methods  of  cal- 
culation in  ordinary  use.  By  means  of  the  Calculus,  how- 
ever, exact  formulas  have  been  deduced,  by  rrieans  of  which 
the  moments  of  inertia  of  many  of  the  simple  geometrical 
forms,  with  respect  to  axes  through  their  centers  of  gravity, 
have  been  found.  There  .are  also  a  number  of  approximate 
methods  by  means  of  which  the  moment  of  inertia  of  an 
irregular  section,  to  which  these  formulas  do  not  apply,  may 
be  found.  Further,  the  moment  of  inertia  of  any  section 
which  can  be  divided  into  parts,  the  moments  of  each  of 
which,  with  respect  to  an  axis  through  its  center  of  gravity, 
can  be  found,  is  easily  calculated  by  means  of  a  principle 
to  be  given  below. 

To  illustrate  the  meaning  of  the  term  moment  of  inertia, 
together  with  a  simple  approximate  method  of  computing 
the  moment  of  inertia  of  a  figure,  consider  the  relation  of 
the  small  I  section  shown  in  Fig.  5  to  the  axis  de  through 
its  center  of  gravity.  Divide  the  section  into  a  number  of 


ARCHITECTURAL  ENGINEERING. 


little  squares  (in  this  case,  each  with  an  area  of  1   square 
inch)  and  consider  the  distance  of  each  square  from  the  axis 


a 

H 

a  a 

a  a   a  a     \. 

i 

>                       t 

i 

7 

i 

' 

»t 

, 

f 

d  rrR^ 
*  '  ^cu  v 

j- 

r 

"-KM    ^ 

•  -  -i 

• 

^Ky    "HOI     1                                         ^ 

*  ]  1—     —  ; 

^ 

• 

i  1 

I 

i-  ^- 

a 

a  a 

a  a  a  a             . 

FIG.  5. 

to  be  the  distance  from  the  axis  to  its  center  of  gravity. 
Then  the  products  of  the  area  of  each  square,  multiplied  by 
its  distance  from  the  axis,  are  as  follows: 

Squares  a,  lx  (-H)2  =  i!i 

Squares  b,  1  X  (4i)'2  =  \J- 

Squares  c ,  1  X  (3i)2  =  ^ 

Squares  d,  1  X  (2^)2  =  ^j5- 
Squares  c,  1  X  (H)2  =  f 
Squares/,  1  X  (|)2  =  1 

Adding  these  products  for  all  the  squares,  we  have : 
16  squares  a,  ifl  X  16  =  4  8  4 

401 


121 


2 
2 
2 
2 
2 

squares  ^, 
squares  c, 
squares  d, 
squares  c, 
squares  / 

Total, 

¥ 

9 

T 

X 
X 
X 
X 
X 

2 
2 
2 
2 

5  G  G 


which  is  the  sum  of  the  products  of  each  of  the  small  areas 
multiplied  by  the  square  of  its  distance  from  the  axis.  This 
result,  however,  is  only  a  rough  approximation  to  the 
moment  of  inertia,  owing  to  the  fact  that  the  assumed  areas 


10  ARCHITECTURAL  ENGINEERING.  §  6 

are  so  large.  The  actual  value  of  the  moment  of  inertia  of 
the  section,  as  will  be  shown  below,  is  568f. 

7.     Rules  and  Formulas  Tor  Moments  of  Inertia. — 

In  the  table  "Elements  of  Usual  Sections"  are  given  exact 
formulas  for  computing  the  moment  of  inertia  of  such  regular 
figures  as  are  most  often  met  in  the  design  of  structures;  it 
also  gives  approximate  formulas  for  computing  this  factor 
for  common  rolled  sections.  The  tables  of  properties  of 
rolled  sections  published  by  the  rolling  mills  give  accurate 
values  of  the  moment  of  inertia  of  all  the  principal  sections 
used  in  the  construction  of  buildings,  so  that  it  is  not  gen- 
erally necessary  to  make  the  calculations  for  these  sections ; 
the  approximate  formulas  in  the  table  "  Elements  of  Usual 
Sections  "  are,  however,  sometimes  useful  in  making  calcu- 
lations when  the  tables  published  by  the  rolling  mills  are 
not  at  hand. 

Rule. —  To  find  the  moment  of  inertia,  with  respect  to  any 
axis  of  any  figure  whose  moment  of  inertia  with  respect  to  a 
parallel  axis  through  its  center  of  gravity  is  known,  add  its 
moment  of  inertia  with  respect  to  the  axis  through  its  center 
of  gravity  to  the  product  of  its  area  multiplied  by  the  square 
of  the  distance  from  its  center  of  gravity  to  the  required 
axis. 

This  rule  may  be  expressed  by  the  formula 
r  =  I+a  r\  (1.) 

in  which  /'  =  the  required  moment  of  inertia ; 

/   =  moment  of  inertia  of  the  section,  with  respect 
to  the  axis  through  its  center  of  gravity  and 
parallel  to  the  given  axis ; 
a    —  the  area  of  the  figure ; 

r    =  the  distance  from  its  center  of  gravity  to  the 
required  axis. 

The  moment  of  inertia,  with  respect  to  an  axis  through 
its  center  of  gravity,  of  any  section  which  can  be  divided 
into  a  number  of  parts,  the  moments  of  inertia  of  each  of 
which,  with  respect  to  an  axis  through  its  center  of  gravity 


§  6  ARCHITECTURAL  ENGINEERING.  11 

parallel  to  the  given  axis,  is  known,  is  equal  to  the  sum  of 
the  moments  of  inertia  of  its  parts  with  respect  to  the  given 
axis.  Since  the  moment  of  inertia  of  any  figure,  with  respect 
to  any  axis,  is  given  by  the  formula  /'  =  I-\-a  r*,  if  we 
denote  the  sum  of  the  moments  of  the  separate  figures 
making  up  a  section,  with  respect  to  an  axis  through  the 
center  of  gravity  of  that  section,  by  2  F  (in  which  the  Greek 
letter  2,  read  signia,  means  sum  of),  we  have 

7s=S/'  =  ;S(/+rtr2),  (2.) 

which  is  a  general  formula  often  used  to  denote  the  moment 
of  inertia  7S  of  any  built-up  section. 

8.  Graphical  Methods  of  Computing?  Moments  of 
Inertia. — There  are  several  graphical  methods  of  compu- 
ting the  moment  of  inertia,  one  of  which  is  an  extension  of 
the  graphical  method  of  locating-  the  center  of  gravity  and 
neutral  axis,  which  was  described  in  Art.  5,  and-  illus- 
trated by  Fig.  4.  Thus,  let  it  be  required  to  determine  the 
moment  of  inertia,  with  respect  to  the  axis  dc  of  the  beam 
section  shown  in  Fig.  4.  Using  the  same  scale  as  that  to 
which  the  section  was  drawn,  compute  or  measure  the  area  of 
the  figure  enclosed  by  the  lines  nopq  .  .  .  .  vivu;  multiply 
this  area  by  the  area  of  the  section — shown  graphically 
by  the  length  of  the  line  a  I — and  the  product  will  be  the 
moment  of  inertia  of  the  section,  with  respect  to  the  axis  de 
through  its  center  of  gravity.  For  example,  suppose  that 
the  section  shown  in  Fig.  4  has  been  drawn  to  a  scale  of 
\  inch  =  1  inch.  Using  this  scale,  and  computing  the  area 
of  the  figure  enclosed  by  the  lines  nop  q  ....  vwn,  we  find 
it  to  be  43.36  square  inches.  The  area  of  the  section  is  01 
square  inches;  therefore,  according  to  the  rule,  its  moment, 
with  respect  to  the  axis  de,  is  43.36x01  =  2,644.06. 

For  finding  the  moment  of  inertia,  it  is  necessary  to 
divide  the  section  into  a  number  of  parts,  for  it  is  evident  that 
the  area  of  the  figure  enclosed  by  the  lines  ;/  r  v  w  n  is  con- 
siderably greater  than  that  of  the  figure  nopq  .  .  .  .  v  w  n, 
obtained  by  dividing  the  web  into  the  small  sections  and 
drawing  the  lines  of  the  diagram  for  each. 


12  ARCHITECTURAL  ENGINEERING.  §  6 

The  area  of  the  figure  n  op  q  .  .  .  .  vwn  may  be  computed 
by  extending  the  horizontal  lines  through  op  q,  etc. ,  so  as  to 
divide  it  into  a  series  of  triangles  and  trapezoids.  The 
dimensions  of  these  can  be  readily  measured,  and  their  areas 
can  be  calculated  by  means  of  the  principles  of  mensuration. 

This  method  of  computing  the  moment  of  inertia  will  be 
found  convenient  in  the  case  of  very  irregular  sections,  to 
which  the  methods  previously  given  can  be  applied  only 
with  considerable  difficulty.  The  accuracy  of  the  result  will 
in  general  be  greater  when  the  section  is  drawn  to  a  large 
scale  and  divided  into  a  comparatively  large  number  of 
parts. 

EXAMPLE  1. — What  is  the  moment  of  inertia  of  the  section  of  a 
10"  X  16"  beam  about  an  axis  through  its  center  of  gravity  parallel  to 
its  shorter  side  ? 

SOLUTION. — From  the  table  "Elements  of  Usual  Sections,"  the 
formula  for  the  moment  of  inertia  of  a  solid  rectangle  is 

bd* 

Substituting  the  given  dimensions,  we  have 

/  *=  — ^ —  =  3,413i.     Ans. 

EXAMPLE  2. — Using  the  approximate  formula  given  in  the  table 
"Elements  of  Usual  Sections,"  compute  the  moment  of  inertia  of  a 
section  of  a  10-inch  I  beam,  the  area  of  the  section  being  10.29  square 
inches. 

SOLUTION. — Substituting  in  the  formula,  we  have 
Af?       10.29X10' 


EXAMPLE  3. — Compute  the  moment  of  inertia,  with  respect  to  the 
axis  de  through  its  center  of  gravity,  of  the  section  shown  in  Fig.  5. 

SOLUTION. — This  section  is  made  up  of  3  rectangles,  the  moments  of 
inertia  of  which,  with  respect  to  the  given  axis,  can  be  found  by  means 
of  formula  1.  The  moment  of  inertia  of  one  of  the  flanges,  with 
respect  to  an  axis  through  its  center  of  gravity  parallel  to  de,  is 

8  XI3 


/  = 


12 


The  area  of  this  figure  is  8  X  1  =  8  square  inches,  and  the  distance 
of  its  center  of  gravity  from  de  is  5|  inches;  therefore,  its  moment  of 


§  G  ARCHITECTURAL  ENGINEERING.  13 

inertia,  with  respect  to  de,  is  /'  =  f  +  $X(5^)2  =  242f.  The  axis 
through  the  center  of  gravity  of  the  web  section  coincides  with  the  axis 
de,  hence  the  moment  of  inertia  of  this  section,  with  respect  to  dc,  is 

1  v  103 

^W~  =  ^ 

Then,  the  moment  of  inertia  /,  of  the  whole  section  =  27'=  242| 
+-  242|  -f-  83i  =  568f.     Ans. 

EXAMPLE  4.  —  What  is  the  moment  of  inertia,  with  respect  to  the  axis 
de,  of  the  column  section  shown  in  Fig.  6  ? 

SOLUTION.  —  -The  moment  of 
inertia  of  one  of  the  flange 
plates,  with  respect  to  an  axis 
through  its  center  of  gravity, 
parallel  to  the  axis  d  e,  is 


The  area  of  the  plate  is  12  X  f 
=  4.5  square  inches,  and  the 
distance  of  its  center  of  gravity 
from  the  axis  is  6T3S  inches. 
Therefore,  the  moment  of  inertia  FIG.  6. 

of  the  plate,  with  respect  to  the  axis  de,  is 

.05  +  4.5  X  (6T\)2  =  172.33. 

From  the  table  "Areas  of  Angles,"  the  area  of  a  4"  X  4"  X  V  angle 
is  3.75  square  inches,  and  the  distance  of  its  center  of  gravity  from 
the  back  of  a  flange  is  approximately  1.25  inches.  The  distance  of  the 
center  of  gravity  of  the  angle  from  the  axis  de,  in  accordance  with 
the  dimensions  given  in  the  figure,  is  6  —  \\  =  4|  inches.  In  accordance 

A  /i* 
with  the  approximate  formula  given  in  the  table  "Elements  of 

Usual  Sections,"  for  finding  the  moment  of  inertia  of  an  angle  with 
equal  legs,  the  moment  of  inertia  of  the  4"  X  4"  X  \"  angle,  with 
respect  to  the  axis  through  its  center  of  gravity,  is 

3.75  X  42 


10.2 


=  5.9,  nearly. 


The  moment  of  inertia  of  the  angle,  with  respect  to  de,  is,  therefore, 
5.9  +  3.75X(4£)2  =  90.5.  The  center  of  gravity  of  the  web-plate  lies 
on  the  axis</^;  therefore,  the  moment  of  inertia  of  the  plate,  with 
respect  to  de,  is 


The  moment  of  inertia  of  the  whole  section,  with  respect  to  de,  is, 
therefore,  2  X  172.  33  +  4  X  90.  5  +  54  =  760.  66.     Ans. 


ARCHITECTURAL  ENGINEERING 


6 


EXAMPLE  5. — What  is  the  moment  of  inertia,  with  respect  to  the 
axis  de,  of  the  column  section  shown  in  Fig.  7  ? 

SOLUTION. — The  moment  of  inertia  of 
one  of  the  cover-plates,  with  respect  to  an 
axis  through  its  center  of  gravity,  paral- 
,n"rL          «  lei  to  de,  is 

/O  Channel  16.5 

*perfoot.  J<J  \s 


The  area  of  the  plate  is  12  X  \  —  6 
square  inches,    and  the   distance  of  its 
FlG-  7-  center  of  gravity  from  de  is   5£   inches, 

therefore,  its  moment  of  inertia,  with  respect  to  de,  is  /'  =  .  125  +  6 
X  (5J)1  =  165.5.  From  Table  12,  Art.  1O7,  Architectural  Engineer- 
ing, §  5,  the  area  of  a  10-inch  16?, -pound  channel  is  4.84  square  inches, 
and  its  moment  of  inertia,  with  respect  to  an  axis  through  its  center 
of  gravity,  corresponding  in  this  case  with  the  axis  de,  is  71.09;  there- 
fore, the  moment  of  inertia  of  the  whole  section,  with  respect  to  d  e,  is 
2X165.54-2X71.09  =  473.18.  Ans. 


RESISTING    MOMENT. 

9.  In  Art.   99,  Architectural  Engineering,   §5,  it  was 
stated  that  the  moment  of  resistance  of  a  section  is  equal  to 
the  product  of  the  greatest  unit  stress  in   any  part  of  the 
section  multiplied  b'y  a  factor  called  the  section  modulus.     In 
calculating  the  strength  of  beams,  it  is,  therefore,  important 
to  be  able  to  determine  the  value  of  the  section  modulus; 
this  can  be  readily  done  for  any  section  whose  moment  of 
inertia  is  known  by  means  of  the  form  ill  a 

K=L,          (3.) 

where  K  —  the  section  modulus; 

/   =  the  moment  of  inertia  with  respect  to  the  neu- 
tral axis; 

and       c    =  the  distance  from  the  neutral  axis  to  the  farthest 
edge  of  the  section. 

1 0.  Relation  Between  Bending  Moment  and  Moment 
of  Resistance. — In  order  that  the  forces  acting  on  a  beam 
may  be  in  equilibrium,  we  know  that  the  moment  of  the 
external  forces,  with  respect  to  a  given  axis,  must  be  equal 
to  the  moments  of  the  stresses  with  respect  to  the  same  axis. 


§  G  ARCHITECTURAL  ENGINEERING.  15 

The  axis  about  which  these  moments  are  assumed  to  act  is 
the  neutral  axis  of  a  section,  and  the  relation  is  expressed  by 
saying  that  the  resisting  moment  must  equal  the  bending 
moment  when  the  beam  is  in  equilibrium. 

Letting  .S"  denote  the  greatest  stress  in  any  fiber  of  a  beam 
subjected  to  a  bonding  stress,  and  M,  the  bending  moment 
in  inch-pounds,  the  relation  between  the  bending  moment 
and  the  resisting  moment  is 

M=  SK  =  S-.  (4.) 

c 

Values  of  the  section  modulus  K  for  the  various  simple 
sections  most  often  used  for  beams  are  given  in  column  2  of 
the  table  "Elements  of  Usual  Sections."  These  values  are 
obtained  by  dividing  the  values  of  /  given  in  column  1  by 
the  distance  c  from  the  neutral  axis  to  the  most  distant  fiber 
of  the  section. 

RADIUS    OF    GYRATION. 

11.  In  computing  the  strength  of  columns,  frequent  use 
is  made  of  a  property  of  a  section  which  is  numerically  equal 
to  the  square  root  of  the  quotient  of  its  moment  of  inertia, 
with  respect  to  an  axis  through  its  center  of  gravity,  divided 
by  its  area.  This  property  is  called  the  radius  of  tf.vra- 
tion  of  the  section,  with  respect  to  the  given  axis.  It  is 
usually  expressed  by  the  letter  A',  and  its  value,  with  respect 
to  a  given  axis,  for  any  section  whose  area  A  and  moment 
of  inertia  /,  with  respect  to  the  same  axis,  are  known,  is 
given  by  the  formula 

R  =  /I'          (5° 

The  form  in  which  the  radius  of  gyration  appears  in  most 
formulas  for  calculating  the  strength  of  columns  is  its  square ; 
hence,  it  is  convenient  to  express  the  above  relation  by  the 
formula 

X*  =  -3,         (6.) 

which  gives  directly  the  value  to  be  substituted  in  the 
column  formulas.  Formulas  for  computing  the  least  radius 
of  gyration  and  its  square  for  the  sections  most  often  used 


1G  ARCHITECTURAL  ENGINEERING.  §  6 

in  the  design  of  structures,  are  given  in  columns  4  and 
5  of  the  table  "Elements  of  Usual  Sections."  The  tables 
of  the  properties  of  rolled  sections  also  give  accurate  values 
of  R  and  R*  for  the  sections  used  in  the  examples  given  in 
the  following  pages. 

EXAMPLE. — Compute  the  square  of  the  radius  of  gyration,  with 
respect  to  the  axis  de,  of  the  column  sectio^i  shown  in  Fig.  6. 

SOLUTION. — By  referring  to  example  4,  Art.  8,  we  find  the  moment 
of  inertia  of  the  section  to  be  /  =  760.66,  and  the  area  of  the  section  is 
3  X  4.5  +  4  X  3.75  =  28.50  square  inches.  Substituting  these  values  in 
formula  0,  we  have 

JP  =  ffiff  =  ML80L     Ans. 


STEEL  COLUMNS. 


TYPES    OF    COLUMNS. 

12.  Columns  of  rolled-steel  shapes  riveted  together  are 
now  largely  used  in  the  construction  of  buildings,  and, 
especially  in  tall  building  construction,  are  rapidly  super- 
seding cast-iron  columns.  On  account  of  their  unreliability 
under  sudden  stress,  cast-iron  columns  are  no  longer  used 
in  bridge  work.  In  buildings,  although  the  loads  are  gener- 
ally quiescent,  and  the  liability  to  sudden  shock  is  more 
remote,  the  columns  seldom  receive  their  loads  as  favorably 
as  in  bridges,  because  in  most  cases  they  are  subjected  to 
considerable  eccentricity  in  loading;  that  is,  the  loads  upon 
one  side  of  a  column  are  heavier  than  those  on  the  other, 
thus  tending  to  produce  bending  stress  and  materially 
decreasing  its  ultimate  strength  to  direct  compression. 

It  may  readily  be  seen  that  the  placing  of  columns  cen- 
trally, one  over  another,  necessitates  the  application  of  loads 
to  their  sides ;  and,  unless  the  loads  are  equal  and  on  opposite 
sides,  the  effect  is  to  increase  the  stress  upon  the  side  where 
the  greatest  load  occurs. 

The  method  of  securing  the  ends  of  columns,  or  struts,  also 
exercises  an  important  influence  on  their  resistance  to  bend- 
ing, and,  consequently,  on  their  ability  to  resist  compression 


6 


ARCHITECTURAL  ENGINEERING. 


17 


stresses.  Columns  may  therefore  be  classified,  according'  to 
the  arrangement  of  their  ends,  into  the  four  following  forms: 
(a)  columns  with  r  <?;///</ ends;  (/>)  columns  with  hinged  ends; 
(c)  columns  with  flat  ends;  and  (</)  columns  with  fixed  ends. 
Fig.  8  shows  the  typical  form  of  each  of  these  classes. 

Round-ended  columns  are  struts  which  have  only  central 
points  or  lines  of  contact,  such  as  balls  or  pins  resting  upon 
flat  plates.  The  centers  of  the  balls  or  pins  should  lie  in  a 
line  through  the  center  of  gravity  of  a  section  of  the  strut. 


Round  Ended 
Cotumn. 


Hinged  Ended  Fieri- Ended 

Co/umn.  Cotumn. 

FIG.  8. 


Fixed  Ended 
Cotumn. 


Columns  with  hinged  ends  are  struts  which  have  both 
ends  properly  fitted  with  either  pins  or  ball-and-socket  joints ; 
the  center  of  the  end  joints  should  lie  on  the  central  axis  of 
the  column. 

Flat-ended  columns  are  struts  which  have  flat  ends  nor- 
mal to  the  central  axis  of  the  strut,  but  not  rigidly  secured 
to  adjoining  parts  of  the  structure. 

1-23 


18 


ARCHITECTURAL  ENGINEERING. 


6 


Columns  with  fixed  ends  are  rigidly  secured  at  both  ends 
to  the  contiguous  parts  of  the  structure  in  such  a  manner, 
that  the  attachment  would  not  be  severed  if  the  member 
was  subjected  to  the  ultimate  load. 

The  round-ended  column  is  seldom  used,  and  in  this 
course  will  be  disregarded  entirely. 

The  column  with  hinged  ends  is  mostly  used  in  pin-con- 
nected trusses,  that  is,  trusses  in  which  the  several  members 
at  a  joint  are  connected  by  means  of  a  pin  or  single  bolt  so 


'"Tens/on  Members 
FIG.  9. 


as  to  be  free  to  turn,  as  on  a  hinge,  instead  of  being  firmly 
riveted,  bolted,  or  spiked  together.  Fig.  9  shows  one  of 
the  lower  joints  of  a  pin-connected  roof  truss. 


FORMULAS    FOR    STEEL,    COLUMNS. 

13.  Columns  AVith  Ilin/ared  Ends. — The  strength  of 
wrought-iron  and  steel  columns  with  hinged  ends,  may  be 
found  from  the  formula 

5  =  1L  (7.) 


L* 

14 . 

r  18,000  J? 


6 


ARCHITECTURAL  ENGINEERING. 


19 


in  which  .S"  =  the  ultimate  breaking  strength  of  the  column 
in  pounds  per  square  inch  of  section ; 

U  =  the  iiltimatc  compressive  strength  of  the 
material  in  pounds  per  square  inch; 

L  =  the  length  of  the  columns  in  inches; 

R  —  the  least  radius  of  gyration  of  the  section. 

This  formula  may  be  expressed  as  a  rule  as  follows : 

Rule. —  To  find  the  ultimate  strength  per  square  inch  of 
sectional  area  of  a  wrought -iron  or  steel  column  with  hinged 
ends,  divide  the  ultimate  compressive  strength  per  square  inch 
of  the  material  composing  the  column  by  1  plus  the  quotient 
obtained  by  dividing  the  square  of  the  length  of  the  column, 
in  inches,  by  18,000  times  the  square  of  the  least  radius  of 
gyration  of  the  section  of  the  column. 

The  safe  bearing  strength  of  the  column  is  obtained  by 
multiplying  its  sectional  area  by  the  ultimate  strength  per 
square  inch,  as  determined  by  formula  7,  and  dividing  this 
product  by  the  factor  of  safety  demanded  by  the  given  con- 
ditions. 

EXAMPLE. — Using  a  factor  of  safety  of  4,  compute  the  safe  bearing 
strength  of  a  strut  of 
a  pin-connected  truss ; 

the  length  of  the  strut  fUJI  <g*6"* &'*"$'«* 

being  20  feet  from  cen- 
ter to  center  of  pins, 
and  its  section  made 
up  of  four  6"  X  6"  X  s" 
angles  connected  as 
shown  in  Fig.  10. 

SOLUTI  ON. — Since 
the  relation  of  the  sec- 
tion to  each  of  the  axes 
a  b  and  d ' e  is  the  same, 
its  moment  of  inertia 
and  radius  of  gyra- 
tion is  the  same  about 
either  axis,  and  it  is  therefore  necessary  to  compute  these  factors  for 
only  one  of  these  axes.  From  the  table  "Areas  of  Angles,"  the  area 
of  the  section  of  a  6"  X  6"  X  f"  angle  is  4.36  square  inches,  and  from 
the  table  "  Properties  of  Angles,"  the  distance  of  its  center  of  gravity 


20  ARCHITECTURAL  ENGINEERING.  §  6 

from  the  back  of  either  leg  is  1.64  inches,  nearly.     By  the  formula 
given  in  the  table  "  Elements  of  Usual  Sections,"  we  find  the  moment 
of  inertia  of  one  of  the  angles,  with  respect  to  an  axis  through  its  center 
of  gravity  parallel  with  the  given  axis,  to  be 
Ah*       4.36X6* 

/=ioa  =   -TojT- 

From  Fig.  10  the  distance  of  the  axis  through  the  center  of  gravity 
of  the  angle  from  the  axis  d  e  of  the  section  is  2.14  inches;  therefore, 
substituting  the  known  values  in  formula  1,  the  moment  of  inertia  of 
a  single  angle,  with  respect  to  the  axis  d  e,  is 

/'  =  f+ar*  =  15.39  +  4.36X2.142  =  35.35. 
The  moment  of  inertia  of  the  whole  section  is,  therefore, 

/„  =  27'  =  4X35.35  =  141.40. 

The  total  area  of  the  section  is  4.36  X  4  =  17.44  square  inches  ;  there- 
fore, the  square  of  its  radius  of  gyration,  from  formula  6,  is 

/?2  =  —  r  =  „.,  '      =  8.1,  nearly. 
A         17.44  * 

In  Table  6,  Art.  61,  Architectural  Engineering,  §  5,  the  ultimate 
compressive  strength  {/of  structural  steel  is  given  as  52,000  pounds  per 
square  inch.  In  accordance  with  the  statement  of  the  problem  the 
length  of  the  strut  is  L  —  20  feet  =  240  inches.  Substituting  in  for- 
mula 7,  we  find  the  ultimate  strength,  per  square  inch  of  section  of  the 
strut,  to  be 

„   >  52,000 

5  =  ~        ~*  -  =  37>27°  P°unds' 


18,000X8.1 
The  safe  bearing  strength  of  the  strut  is,  therefore, 

37,270  X  17.44 

—  j  —      -  =  162,500  pounds,  nearly.  Ans. 

14.  Columns  With  Flat  or  Square  Ends.  —  The  col- 
umns used  for  supporting  the  tiers  of  floorbeams  in  an  office 
building  may  properly  be  considered  as  columns  with  flat  or 
square  ends,  and  their  strength  may  be  calculated  by  the  fol- 
lowing formula  for  flat  or  square  ended  columns  : 


in  which  the  symbols  S,  U,  L,  and  R  have  the  same  mean- 
ing as  in  formula  7. 

EXAMPLE.  —  The  moment  of  inertia,  with  respect  to  the  axis  Y  Y  of 
the  steel  Z-bar  column  shown  in  Fig.  11,  is  287.92,  and,  with  respect  to 


ARCHITECTURAL  ENGINEERING. 


the  axis  XX,  337.17.  The  total  area  of  the  section  is  21.36  square 
inches.  What  is  the  safe  bearing  strength  with  flat  ends,  if  the  length 
of  the  column  is  20  feet  and  a  factor 
of  safety  of  3  is  used  ? 

SOLUTION. — Using  the  least  mo- 
ment of  inertia,  which  is  that  with 
respect  to  the  axis  Y  V,  the  square 
of  the  least  radius  of  gyration  is 
found  to  be 


=  13.48,  nearly, 


21.36 


therefore,  the  ultimate  strength  per  square  inch  of  section  is 
52,000 


S  = 


240- 


=  44,100  pounds. 


1  24,000  X  13.48 

The  safe  bearing  strength  is,  therefore, 
44,100  X  21.36 


3 


=  314,000  pounds,  nearly.     Ans. 


15.  Columns  AVith  Fixed  Ends. — Letting  S,  U,  L,  and  R 
represent  the  same  quantities  as  in  the  two  preceding  for- 
mulas, the  strength  of  columns  with  fixed  ends  may  be 
calculated  by  the  formula 

5  = V-  (9.) 


1  36, 000  A32 

EXAMPLE. — A  section  of  the  compression  member  of  a  large  struc- 
tural-steel roof  truss  is  shown  in  Fig.  12. 
The  ends  of  the  member  are  firmly 
riveted  to  the  adjacent  members  of  the 
truss,  and  the  length  of  the  member  is  15 
feet;  what  is  its  safe  bearing  strength 
with  a  factor  of  safety  of  4  ? 
are:  SOLUTION. — From  the  table  "Areas  of 

e  Angles,"  the  area  of  a  3"  X  3"  X  }"  angle 

is  1.44  square  inches,  and  from  the  table 
"Properties  of  Angles,"  the  distance  of 
III  its  center  of  gravity  from  the  back  of  a 

— ^""^P""1^  flange  is  .84  inch;    also  from  the  latter 

table,  the  moment  of  inertia  of  the  angle, 
with  respect  to  an  axis  through  its  center 
of  gravity,  is  found  to  be   1.24.     In  accordance  with  the  dimensions 


FIG.  12. 


22  ARCHITECTURAL  ENGINEERING.  §  6 

given  in  the  figure,  the  distance  from  the  axis  de  to  the  centers 
of  gravity  of  the  angles  is  4  — .84  =  3.16  inches.  The  moment  of 
inertia  of  one  of  the  angles,  with  respect  to  the  axis  d  e,  is,  therefore, 

/'  =  1.24 +  1.44X3- IS"  =  15.62. 

The  moment  of  inertia  of  the  web-plate,  with  respect  to  d  e,  is 

<•  v  K3 

/'  =  I**  1  -  13.33. 
1* 

The  moment  of  inertia  of  the  whole  section,  with  respect  to  de,  is 

Is  =  4  X  15.62  +  13.33  =  75.81. 

The  total  area  of  the  section  is  4  X  1-44  +  8  X  -fg  =  8.26  square  inches ; 
the  square  of  its  radius  of  gyration,  with  respect  to  de,  is,  therefore, 

75.81 

R    =-aW  =  9'ia 

Referring  now  to  the  axis  a  b,  the  distance  from  the  axis  to  the  cen- 
ter of  gravity  of  one  of  the  angles  is.84  +  /j  =  .84 +  .156  =  .996,  say 
1  inch. 

The  moment  of  inertia  of  the  angle,  with  respect  to  a  b,  is,  therefore, 

/'  =  1.24  + 1.44  X  I2  =  2.68. 
The  moment  of  inertia  of  the  web-plate,  with  respect  to  a  b,  is 

8  X  A3  =    02 
12 

Taking  the  sum- of  the  moments  of  inertia  of  the  several  parts,  the 
moment  of  inertia  of  the  whole  section,  with  respect  to  a  b,  is  found  to 
be  4  X  2.68  +  .02  =  10.74;  dividing  this  by  the  area  of  the  section,  the 
square  of  the  radius  of  gyration,  with  respect  to  this  axis,  is 

r,n  10.74 

"8~^6  =     '  nearly> 

which,  being  much  less  than  the  value  of  R*  when  referred  to  the  axis 
d  e,  is  the  value  to  use  in  computing  the  strength  of  the  member. 

Substituting  now  in  formula  9,  the  ultimate  strength  per  square 
inch  of  the  section  is 

RO  ooo 
*"a          =  30,700  pounds, 


36,000X1.3 
and  the  safe  bearing  strength  of  the  member  is 

30,700X8.26 

— —     -  =  63,400  pounds.     Ans. 

16.     Columns      With      Eccentric      Ijoatls. — Columns 
designed   by   the   foregoing-   formulas   are    reasonably   safe 


ARCHITECTURAL  ENGINEERING. 


against  crushing,  bending,  or  buckling  unless  subjected  to 
considerable  bending  stress,  due  to  eccentric  loading. 

In  modern  buildings  nearly  all  columns  are  subjected  to 
more  or  less  eccentric  loading.  In  fact,  it  would  be  a  diffi- 
cult matter  to  find  a  column  loaded  with  a  perfectly  concen- 
tric load.  It  is,  however,  good  practice  to  disregard  the 
eccentric  loading  generally  encountered  in  ordinary  building 
construction,  since  the 
preceding  f  o  r  m  u  1  a  s 
make  due  allowance  for 
the  slight  eccentricity 
caused  by  the  lack  of 
symmetry  in  the  ar- 
rangement of  the  brack- 
ets and  the  applica- 
tion of  the  loads  with 
regard  to  the  central 
axis  of  the  column.  For 
example,  the  beam  con- 
nections  may  all  be 
upon  one  side  of  the 
column,  and  of  the  form 
shown  in  Fig.  13,  or  the 
bsams  attached  to  one 
side  of  the  column  may 
be  more  heavily  loaded 
than  those  attached  to 
the  other  side.  Some 
of  these  conditions  are 

usually  unavoidable   and   tend  to   produce    the  undesirable 
effect  of  eccentric  loading. 

Should  the  eccentricity  of  the  load  be  considerable,  and 
liable  to  produce  dangerous  transverse  or  bending  stresses, 
it  would  materially  decrease  the  ability  of  the  column  to 
withstand  direct  coinprcssive  stresses.  The  bending  or 
transverse  stresses  should  in  such  a  case  be  calculated,  and 
an  additional  amount  of  material  should  be  added  to  the 
section  of  the  column  in  order  to  resist  them. 


24 


ARCHITECTURAL  ENGINEERING. 


A  column  which  is  a  fair  example  of  eccentric  loading 
is  shown  in  Fig.  14.  The  bending  moment  or  trans- 
verse stress  is  equal  to  the  product 
of  the  load  by  the  lever  arm  through 
which  it  acts,  in  this  case  2  feet. 
Hence,  the  bending  moment  upon 
the  column,  due  to  the  eccentric 
load,  is  10,000X2  =  20,000  foot- 
pounds, or  240,000  inch-pounds.  The 
problem  now  is  to  determine  the 
amount  of  material  required  and 
its  disposition  in  order  to  provide 
sufficient  resistance  to  this  trans- 
verse or  bending  stress;  its  solu- 
tion, however,  depends  on  the 
principles  involved  in  the  design 
of  beams  and  girders,  principles 


j 


FIG.  14. 

that  will  be  more  fully 
discussed  in  connection 
with  the  following  ar- 
ticles. 

The  failure  of  a  struc- 
tural-steel column  may 
occur  by  reason  of  the 
buckling  between  the 
rivets  of  the  plates  or 
structural  shapes,  of 

which  it  is  constructed;  the  homogeneousness  of  the  sec- 
tion will  thus  be  affected,  as  may  be  seen  by  reference  to 
Fig.  15. 


FIG.  15. 


§  G  ARCHITECTURAL  ENGINEERING.  25 

17.  Factors  of  Safety.  —  The  ability  of  a  column 
to  resist  the  transverse  or  bending1  stresses  due  to  eccen- 
tric loading  decreases  as  its  length  increases,  and  this 
ability  is  less  for  columns  with  round  or  hinged  ends 
than  for  those  with  flat  or  fixed  ends.  For  these  rea- 
sons it  is  good  practice  to  assume  a  minimum  factor  of 
safety  for  the  shortest  columns  or  struts  and  increase  the 
factor  with  the  increase  in  length,  making  the  rate  of 
increase  greater  for  round  or  hinged  than  for  flat  or  fixed 
ends. 

Assuming  a  minimum  factor  of  safety  of  3  for  very 
short  struts,  the  factors  prescribed  by  good  practice  for 
longer  struts  with  flat  or  fixed  ends  are  given  by  the 
formula 

I~;  (10.) 


and  for  struts  with  round  or  hinged  ends, 

F==  3  +  .015  ^  (11.) 

In  both  of  these  formulas  F  is  the  required  factor  of 
safety;  /,  the  length  of  the  strut  in  inches;  and  R,  the  least 
radius  of  gyration  of  the  section. 

EXAMPLE.  —  What  is  the  least  factor  of  safety  that  should  be  used 
with  (a)  a  column  with  flat  ends,  and  (b)  a  column  with  hinged  ends, 
the  length  of  the  column  being  20  feet,  and  its  least  radius  of  gyration 
2.5? 

SOLUTION.  —  (a)  Applying  formula  1O,  we  have 

240 

F  =  3  +  .01  X  TT=  =  3.96,  say  4.     Ans. 
&.  o 

(b)  Applying  formula  11, 

240 

F  =  3  +  .015  X  Ins  =  4-44-     Ans- 
2.5 

Table  1  gives  values  of  the  factor  of  safety  obtained 
by  substituting  different  values  of  „  in  formulas  1O 
and  11. 


26 


ARCHITECTURAL  ENGINEERING. 


TABLE    1. 

FACTORS  OF  SAFETY. 


Fixed 
and 

Hinged 
and 

/ 

Fixed 
and 

Hinged 
and 

/ 

Fixed 
and 

Hinged 
and 

A1 

Flat 

Round 

A' 

Flat 

Round 

A 

Flat 

Round 

Ends. 

Ends. 

Ends. 

Ends. 

Ends. 

Ends. 

20 

3.2 

3.30 

110 

4.1 

4.G5 

200 

5.0 

6.00 

30 

3.3 

3.45 

120 

4.2 

4.80 

210 

5.1 

6.15 

40 

3.4 

3.  GO 

130 

4.3 

4.95 

220 

5.2 

6.30 

50 

3.5 

3.75 

140 

4.4 

5.10 

230 

5.3 

6.45 

GO 

3.G 

3.90 

150 

4.5 

5.25 

240 

5.4 

6.60 

70 

3.7 

4.05 

1GO 

4.G 

5.40 

250 

5.5 

6.75 

80 

3.8 

4.20 

170 

4.7 

5.55 

260 

5.6 

6.90 

90 

3.9 

4.35 

180 

4.8 

5.70 

270 

5.7 

7.05 

100 

4.0 

4.50 

190 

4.9 

5.85 

280 

5.8 

7.20 

The  table  gives  values  of  the  factor  of  safety  for  columns 
whose  length  is  as  great  as  280  times  their  least  radius  of 
gyration ;  it  is  nqt,  however,  good  practice  to  use  a  column 

in  which  the  value  of  -^  is  greater  than  150.  Another  con- 
dition to  be  observed  is  that  the  length  of  the  column  should 
not  exceed  45  times  its  least  dimension. 


FORMS    OF    STEEL    COLUMNS. 

18.  Conditions  AVhich  Affect  the  Choice  of  a  Type 
of  Column. — There  are  at  present  numerous  forms  in 
which  rolled-steel  shapes  are  combined  to  make  up  columns 
for  structural  purposes.  The  type  of  column  to  be  used  in  a 
building  is  sometimes  prescribed  by  the  owner,  or  the 
design  of  the  building  is  furnished  by  the  architect,  thus 
leaving  the  engineer  little  latitude  in  his  choice  of  design. 
There  are,  however,  a  number  of  conditions  demanded  by 
considerations,  partly  practical  and  partly  theoretical,  which 
should  be  carefully  studied  and  compared  before  selecting 


§0  ARCHITECTURAL  ENGINEERING.  2? 

the  type  of  column  for  a  particular  purpose.  In  many  eases 
it  will  be  found  that  these  considerations  impose  conditions 
that  conflict  with  each  other;  and  in  order  to  make  such  a 
compromise  as  will  meet  this  difficulty  in  the  most  satisfactory 
manner,  there  is  demanded  of  the  engineer  a  most  careful 
exercise  of  judgment  guided  by  practical  experience.  Some 
of  the  most  important  points  to  be  considered  in  choosing 
the  type  of  column  to  be  used,  in  any  case,  are  the  following: 

1.     The  cost  and  availability  of  the  material. 

"Z.  The  amount  of  labor  required  and  the  facility  with 
which  it  may  be  performed  in  both  shop  and  field. 

3.  The  distribution  of  material  in  the  column  so  as  to 
give  the  maximum  strength  with  the  least  weight. 

4.  The   facility   with   which   connections    may   be    made 
between  the  column  and  the  members  which  it  supports. 

5.  The  application  of  the  connections  in  such  a  manner 
that  they  will  transfer  the  compressive  stresses  directly  to 
the  axis  of  the  column. 

0.  The  facility  with  which  the  thickness  of  the  metal  in 
the  different  parts  composing  the  column  can  be  reduced  in 
order  to  meet  the  reduced  loads  of  the  upper  floors.  It  is 
not  desirable  to  make  the  columns  supporting  the  upper 
floors  of  the  building  very  small,  since  the  beams  and 
girders  supporting  the  upper  floors  are  usually  of  the  same 
dimensions  as  those  for  the  lower  floors,  and  consequently 
require  connections  as  heavy  and  secure;  it  is  almost 
impossible  to  make  such  connections  to  small  columns, 
consequently,  in  order  to  reduce  the  weight  of  the  column  for 
the  lighter  load  which  it  will  carry,  it  is  better  to  reduce  the 
thickness  of  the  material  used  and  keep  the  section  the  same. 

7.  The  facility  with  which  fireproofing  may  be  attached 
to  the  section.  Columns  of  circular  sections  may  be 
fireproofed  more  compactly  than  rectangular  columns. 
Architects  in  some  instances  have  utilized  the  space  lost  by 
the  use  of  rectangular  columns  for  the  purpose  of  conduits 
in  which  to  run  pipe  lines,  etc.  In  one  instance,  for  example, 


28  ARCHITECTURAL  ENGINEERING.  §  6 

circular  holes  were  cut  in  the  bed  and  cap  plates  between 
the  columns  of  several  stories,  to  allow  for  the  passage 
of  pipes  inside  of  the  column.  Such  practice  cannot  be 
condemned  too  severely.  It  is  bad  practice  to  run  pipe  lines 
or  electric  wires  inside  of,  or  alongside  of,  the  fireproofing 
of  columns ;  separate  conduits,  provided  with  suitable  covers, 
which  will  allow  of  inspection,  should  be  provided  in  the  walls. 

In  regard  to  the  cost  and  availability  of  the  material,  such 
shapes  shoiild  be  selected  as  are  easily  rolled  and  placed  on 
the  market  at  a  reasonable  price.  I  beams,  channels,  angles, 
and  Z  bars,  together  with  plates,  are  the  most  common 
commercial  sections ;  these  are  manufactured  at  nearly  every 
structural-steel  mill,  and  may  be  obtained  promptly  on  large 
orders  at  any  locality  where  a  skeleton-constructed  building 
is  likely  to  be  erected.  Patented  sections  do  not  fulfil  this 
condition,  and  should  therefore  be  avoided  in  most  cases. 
The  consideration  of  prompt  delivery  is  an  important  one, 
and  greatly  influences  the  cost  and  facility  with  which  the 
modern  building  is  erected. 

The  consideration  of  the  facility  with  which  the  labor  in 
both  shop  and  field  can  be  performed  is  one  that  should 
receive  careful  attention.  In  the  shop  the  complexity  of 
the  column  section,  and  the  number  of  pieces  of  which  it  is 
composed,  greatly  influence  the  element  of  labor.  If  there 
are  numerous  small  pieces,  such  as  brackets  or  splice  plates, 
each  of  which  requires  cutting,  bending,  and  fitting  together, 
with  frequent  handling,  the  cost  of  labor  may  be  proportion- 
ately very  great.  The  number  of  rivets  also  greatly 
influences  the  cost  of  a  column ;  not  only  should  there  be  as 
few  rivets  as  is  consistent  with  strength,  but  the  construc- 
tion should  permit  the  rivets  to  be  readily  driven  by  machine, 
so  as  to  avoid  hand  riveting,  which  is  slow  and  expensive 
and  now  generally  admitted  to  be  inferior  to  machine  work. 

The  same  general  remarks  apply  to  the  labor  in  the  field ; 
the  connections  should  be  as  simple  as  possible,  and  with  as 
few  rivets  as  is  consistent  with  the  strength  required ;  they 
should  be  easy  of  access,  so  that  the  work  upon  them  may 
be  executed  conveniently  and  rapidly. 


§  6  ARCHITECTURAL  ENGINEERING.  29 

In  connection  with  the  question  of  the  distribution  of  the 
material  in  the  column  so  as  to  develop  the  maximum 
amount  of  strength  with  the  least  weight  of  material,  it  is 
necessary  to  consider  the  facility  with  which  economical 
connections  may  be  made  between  the  columns  and  floor- 
beams,  and  the  directness  with  which  the  connections  trans- 
fer the  stresses  to  the  central  axis  of  the  column.  The 
relation  between  these  conditions  may  be  illustrated  and 
analyzed  by  comparing  the  sections  shown  in  Fig.  10,  where 


Floor  Beams. 


FIG.  16. 

(a)  represents  a  section  of  a  Phoenix  column,  while  (/>)  is  a 
section  of  a  Z-bar  column.  The  apparent  advantage  of  the 
Phoenix  column  is  that  the  material  composing  it  is  placed 
where  it  will  be  the  most  efficient,  thus  fulfilling  the  third 
condition  of  the  above  list  in  a  satisfactory  manner;  the 
radius  of  gyration  of  the  column  is  also  practically  the  same 
upon  any  of  its  diameters,  and  the  metal  is  placed  at  the 
greatest  possible  distance  from  its  center. 

On  the  other  hand,  by  referring  to  the  section  of  the  Z-bar 
column,  it  is  seen  that  a  considerable  portion  of  the  mate- 
rial is  concentrated  on  the  axis,  a  condition  that  gives  it  a 
relatively  small  radius  of  gyration  and  demands  the  use  of  a 
greater  weight  of  material  for  a  given  load.  This  column, 


30 


ARCHITECTURAL  ENGINEERING. 


6. 


however,  offers  greater  advantages  for   the  proper  connec- 
tion of  the  floorbeams  than  does  the  other,  and,  owing  to  its 
open  construction,  the  beams   transmit   their  loads  almost 
directly  to  the  central  axis  of  the  col- 
umn,  thus   avoiding    the    disadvan- 
tages of  eccentric  loading. 

Thus  it  is  seen  that  the  sec- 
tion having  the  best  theoretical 
distribution  of  material  is  not 
always  the  best  to  use,  on  account 
of  these  several  practical  considera- 
tions; in  fact,  the  section  shown  in 
Fig.  17  is  probably  as  much  used 
for  structural-steel  columns  as  any 
other,  although  it  is  not  an  econ- 
omical section,  for  the  reason  that 
its  radius  of  gyration  on  the  axis 
dc  is .  very  small  in  proportion  to 
the  weight  of  material  used,  and, 
since  the  least  radius  of  gyration 
is  always  used  in  calculating  the 
strength  of  a  structural  column,  all 
the  material  which  goes  to  form  the 
greater  radius  of  gyration  adds  noth- 
ing to  the  theoretical  strength. 

The  great  advantage  of  this  section 
is  that  it  is  composed  of  the  cheap- 
est rolled  sections,  which  are  put 
together  with  a  minimum  of  labor. 
It  is  also  one  of  the  best  forms  for 
attaching  the  beams  and  girders,  and,  as  will  be  seen  by  a 
study  of  the  details  of  the  splice  shown  in  the  figure,  for 
making  connections  between  two  adjacent  columns. 

19.  Sections  of  Columns  Frequently  Used  in  Skele- 
ton Building  Construction. — The  following  list  shows  the 
general  forms  of  the  principal  types  of  steel-column  sections 
now  in  use: 


rp*  —  ^ 

^  

9 

9 

9    9          i 

1 

9 

9 

9  9       ; 

1 

9 

9 

!' 

9    9 

, 

9 

>Q 

9    9           1 

Q 

*        9 

9^9           i 

1 

9 

^         L® 

9 

~0l 

*        9 

9\9       ; 

jj 

9 

*       9 

9    9          i 

i 

9 

9 

(. 

9    9          1 

9 

9 

[ 

9    9          I 

r 

9 

9 

9  9       ; 

>  ; 

9 

9 

k 
. 

_/^x~^. 

pi 

, 

; 
-^ 

Fio. 


§  6  ARCHITECTURAL  ENGINEERING. 


[—4  er-|  Larimer  column,  1  row  of  rivets  (patented). 


Z-bar  column  without  cover-plates,  2  rows 
of  rivets. 


f 

-*• 

— 

— 

•?- 

•*• 

— 

— 

Z-bar  column  with  one  cover-plate,  G  rows 
of  rivets. 


Z-bar  column  with  two  cover-plates, G  rows 
of  rivets. 


Z-bar  column,  additional  section  obtained 
by  the  use  of  angles  and  plates,  8  rows  of 
rivets. 


Z-bar  column,  rectangular  section,  G  rows 
of  rivets. 


Channel  column  with  plates  or  lattice,  4 
rows  of  rivets. 


Box  column  of  plates  and  angles,  8  rows 
of  rivets. 


32  ARCHITECTURAL  ENGINEERING.  §  6 


Latticed  angle  column,  8  rows  of  rivets. 


A  column  much  used  by  the  Pennsylvania 
Railroad,  composed  of  angles  and  plates,  10 
r°ws  of  rivets. 


Keystone    octagonal   column,    4    rows   of 
rivets  (patented). 


••  Four-section   Phoenix   column,  4   rows  of 
rivets  (patent  expired). 


Eight-section  Phoenix  column,  8  rows  of 
rivets  (patent  expired). 


Grey  column,  4  rows  of  rivets  (patented). 


§  6  ARCHITECTURAL  ENGINEERING.  33 

DETAILING    OF    STRUCTURAL    COLUMNS    AND 
CONNKCTIONS. 

2O.  Rivets  and  Rivet  Spacing. — Before  discussing"  the 
design  of  structural-steel  columns  and  their  connections, 
we  will  consider  the  best  practice  in  proportioning  and 
spacing  the  rivets  which  connect  the  several  rolled  sections 
of  which  the  completed  column  is  composed. 

Where  brackets  are  riveted  to  the  column  to  support  floor- 
beams,  a  sufficient  number  of  rivets  should  be  used  to  take 
the  entire  shear  due  to  the  reaction  at  the  end  of  the  beam ; 
also  where  a  plate  girder,  which  forms  the  principal  member 
of  a  floor  system,  is  riveted  directly  to  the  column,  sufficient 
rivets  should  be  provided  so  that  their  combined  shearing 
strength  is  equal  to  the  end  reaction  on  the  girder.  If,  as 
shown  in  Fig.  13,  there  is  a  bracket  or  knee  brace  in  connection 
with  the  girder,  the  calculation  may  be  based  on  the  combined 
shearing  strength  of  the  rivets  in  both  bracket  and  girder. 

EXAMPLE. — If  the  reaction  at  the  end  of  a  plate  girder  is  60,000 
pounds,  and  |-inch  rivets,  each  with  an  allowable  shearing  strength  of 
6,500  pounds,  are  used,  how  many  rivets  will  be  required  to  support  the 
end  of  the  girder  ? 

SOLUTION.—    60,000  Ib.  -5-6,500  Ib.  =  9.2,  say  10  rivets.     Ans. 

It  is  usual  to  space  the  rivets  closer  at  the  joints  and  foot 
of  a  column  than  in  the  body;  for  example,  where  f-inch 
rivets  are  used,  it  is  customary  to  space  them  on  3-inch 
centers  at  the  joints  and  bottom  and  from  44  to  G  inch 
centers  throughout  the  remainder  of  the  column,  and  where 
| -inch  rivets  are  used,  they  should  be  spaced  about  4  inches 
at  the  joints  and  about 


u  (,    ^Pitch  of  these  rivets  5 

6    inches  throughout    the  ^ 
length  of  the  column. 


In  a  compression  mem- 
ber the  pitch  of  the  rivets 
in  inches  should  never  ex- 
ceed the  thickness  in  IGths 
of  an  inch  of  the  thinnest  FIG-  ia 

outside  plate.      For  example,  in  Fig.  18,  where  the  outside 

1-24 


F 


Outside  P/ffte. 


34  ARCHITECTURAL  ENGINEERING.  §  6 

plate  is  T5F  inch  thick,  the  greatest  allowable  pitch  of  the 
rivets  would  be  5  inches;  under  no  conditions,  however, 
should  the  pitch  of  the  rivets  in  a  compression  member  be 
greater  than  6  inches,  center  to  center,  except  where  the 
rivets  are  staggered,  in  which  case  the  pitch  should  not  be 
more  than  6  inches  in  a  staggered  line. 

The  diameter  of  rivets  to  be  used  in  built-up  columns,  is 
determined  by  the  thickness  of  the  metal  in  the  parts  to  be 
joined.  Where  the  aggregate  thickness  of  the  metal  between 
the  heads  of  the  rivet  is  not  more  than  1^  inches,  a  f-inch 
rivet  may  be  used ;  if  the  aggregate  thickness  of  the  several 
plates  is  more  than  1^  inches  and  less  than  3  inches,  it 
would  be  advisable  to  use  |--inch  rivets;  and  if  the  aggregate 
thickness  of  the  metal  is  3  inches  or  more,  1-inch  rivets 
should  be  used. 

As  far  as  practicable  the  rivets  should  be  of  the  same  size 
throughout  the  column,  as  this  saves  annoyance  in  both  shop 
and  field.  The  size  of  the  rivets  to  be  used  in  the  column 
splices,  girder  and  beam  connections,  knee  braces,  and 
brackets  is  determined  by  the  size  of  the  rivets  in  the 
column. 

In  conjunction  with  the  above  rules  the  student  should 
exercise  his  judgment  in  the  matter  of  details,  and  should 
study  the  requirements  of  each  condition  to  be  fulfilled. 

21.  Column  Splices  and  Connections. — The  excel- 
lence of  the  detail  design  of  structural-steel  columns  is 
largely  governed  by  the  experience  and  judgment  of  the 
designer,  and  the  care  with  which  he  studies  the  local  con- 
ditions which  will  be  met  with  in  nearly  every  new  piece  of 
work. 

The  several  points  of  excellence  to  be  attained  in  the 
design  of  structural  cohtmns  have  already  been  discussed, 
and  it  now  remains  for  the  student  to  consider  a  few  first- 
class  details  of  column  splices  and  connections. 

As  the  strength  of  a  building  depends  almost  entirely 
upon  the  strength  of  the  connections  to  the  columns, 
great  care  should  be  taken  in  their  design.  Rigid  column 


6 


ARCHITECTURAL  ENGINEERING. 


connections  for  the  floorbeams  andgirders  at  the  several  floors, 

and  efficient  splice  connections  between  the  .sections  of  the 

columns  on  the  different  floors  are  of  the 

utmost  importance,  and  should  be  given  the 

most  careful  attention.     The  ideal  system 

of  column  construction  would  be  to  make 

the  various  columns  on  the  several  floors 

of  one  continuous  set  of  sections  running 

from    foundation    to   roof;    it   is   evident, 

however,  that  this  is  impossible  in  high 
modern  building  construc- 
tion, therefore  the  next 
best  thing  to  do  is  to  make 
the  spliced  connections  as 
03  rigid  as  possible. 

Columns  may  be  spliced 


o! 


FIG.  20. 


as  shown  in  Fig.  19.  The 
bedplate  a  separates  the 
two  columns,  and  through 
it,  by  means  of  the  angles 
/>,  //>,  the  columns  are  rivet- 
ed securely  together;  they 
are  also  additionally  se- 
cured by  means  of  the 
two  splice  plates  c,  c. 

The  abutting  ends  of 
structural  columns  should 
be  milled  or  planed  so  as 
to  secure  a  square  and 
firm  bearing  and  obviate  the  danger  of  their 
being  thrown  out  of  line  when  erected. 

Another  very  good  method  of  connecting 
two  columns  is  by  the  use  of  splice  plates 
on  all  sides,  thus  doing  away  with  the  bed- 
plates or  packing  pieces  between  the  ends ; 
construction  is  shown  in  Fig.  20.  The 


Fl°- 


bedplate    is   sometimes   extended  beyond    the  column  and 
forms  a  rest  or  support  for  the  floorbeams  or  girders.      If 


36 


ARCHITECTURAL  ENGINEERING. 


6 


this  is  done,  it  is  advisable  to  further  support  the  bedplate 
by  a  bracket  directly  under  the  bearing-  of  the  beams,  as 
shown  in  Fig.  21. 

If,  in  the  same  floor,  the  beams  or  girders  are  of  different 


FIG.  21. 

depths  and  are  at  the  same  level  with  regard  to  the  top 
flange,  the  shallow  beams  may  be  supported  by  introducing 
cast-iron  blocks  beneath  them,  as  shown  at  a,  Fig.  21. 

Where  the  floor  girders  are  of  the  built-up  plate-girder 
type,  there  is  no  difficulty  in  securing  a  rigid  connection 


6 


ARCHITECTURAL  ENGINEERING. 


37 


to  the  column.      The  connection  to  be 
architectural    features   of 
the  interior  arrangement 
of  the  building  do  not  in- 
terfere is  shown  in  Fig.  22. 
This'connection  is  partic- 
ularly efficient  when'  the 
building  is  high  and  nar- 
row,  and  in    danger 
of  being  acted  on  by 
heavy  wind  pressure. 
Other    forms    of 
connections,     where 
plate  girders  are  the 
principal  supporting 
members  of    a  floor 
system,  are  shown  in 
Figs.  23  and  24.     In 


recommended  where  the 


^    o    v    v    t 

t 

^^"oVt 

\ 

c 

C 

\ 

4 

/ 

t 

C 

1 

1 

\ 

© 

c 

) 

© 

S 

f 

0 

c 

r©  ©  ©  ©  © 

C        4,         ^.      i.       L.     V 

_^.  — 

m 

%—  - 

1  —  ^I1 

c 

© 

1  c 

© 

^ 

© 

a 

< 

i    JO 
i    ? 

© 

<  >© 

1 

« 

o 

5 

© 

»! 

l,o  o  ©J 

[5  5  ©| 

i  f  c 

\\ 

D© 

S 

©j 

c 

© 

,. 

c 

© 

© 

i 

© 

© 

© 

© 

< 

© 

f*^ 

o 

"vll 

FIG.  23. 


FIG.  22. 

Fig.  23,  it  will  be 
seen  that  a  filler  is 
introduced  at  c  be- 
cause the  upper  col- 
umn is  smaller  than 
the  lower  and  it  be- 
comes necessary  to 
pack  between  the 
splice  plates.  It  is, 
however,  preferable 
to  pack  equally  on 
both  sides  of  the 


38 


ARCHITECTURAL  ENGINEERING. 


6 


upper  column  whenever  practicable,  as  this  insures  the 
central  axis  of  the  one  column  being  over  the  central  axis  of 
the  other. 


Q    Q     Q    Q 


|o>    Q    Q 

\ 

Q 
O 

i  i         i  nl 

1 

1 

Q      Q     Q 

/ 

FIG.  24. 

22.  Examples  of  Good  Column  Design. — In  Figs.  25 
and  26  are  shown  complete  detail  working  drawings  of  the 
first-floor  columns  of  a  building  designed  according  to  the 
best  modern  practice. 

Fig.  25  is  a  channel  column.  The  floorbeams  are  rigidly 
secured  to  the  column  at  a  considerable  distance  below  the 


(i  ARCHITECTURAL  ENGINEERING. 


39 


FIG.  25. 


40 


ARCHITECTURAL  ENGINEERING. 


~B  Sectional  Plan  onAB 


All  Rivets  £  diam.unlesa 
otherwise  marAecf. 


-31k 


FIG.  28. 


§  G  ARCHITECTURAL  ENGINEERING.  41 

splice,  which  has  been  designed  with  a  bedplate  between  the 
two  sections.  The  splice  could  have  been  made  equally  as 
well,  and  would  possibly  have  been  more  efficient,  by  the 
use  of  side-splice  plates  as  previously  explained. 

In  Fig.  20  is  shown  a  Z-bar  column,  the  principal  points 
in  the  detail  of  which  are  similar  to  those  in  the  channel-bar 
column. 

In  each  of  the  above  examples,  the  shear  on  the  rivets  sup- 
porting the  floorbeam  brackets  has  been  calculated  to  safely 
sustain  the  load  carried  upon  the  ends  of  the  floorbeams. 

All  the  connections  and  splices  to  a  column  should  be 
riveted  together  with  hot  rivets;  this  insures  more  rigidity 
against  wind  pressure  than  can  be  obtained  by  bolted 
connections. 

23.  Rivet  Signs. — In  making  drawings  of  structural- 
steel  work,  the  form  of  rivet  to  be  used  should  be  designated 
by  some  system  of  conventional  signs.  The  accompanying 
list  shows  Osborn's  code  of  conventional  signs  for  rivets, 
which  has  been  adopted  by  many  of  the  leading  bridge 
companies  and  consulting  engineers,  and  by  many  of  the 
railway  companies. 

The  basis  of  this  system  consists  of  the  open  circle  to  rep- 
resent a  shop  rivet  and  the  blackened  circle  to  represent  a  field 
rivet,  the  diagonal  cross  to  indicate  a  countersunk  head  and 
the  vertical  stroke  to  indicate  a  flattened  head.  The  position 
of  the  diagonal  lines  with  reference  to  the  circle  (inside,  out- 
side, or  both)  indicates  whether  the  rivet  head  is  counter- 
sunk into  the  inside,  outside,  or  both  sides  of  the  material. 
Similarly,  the  number  and  position  of  the  vertical  strokes 
indicate  the  height  and  position  of  the  flattened  head.  Any 
combination  of  shop  or  field  rivets  with  full,  countersunk,  or 
flattened  heads,  maybe  readily  indicated  by  the  proper  com- 
bination of  these  signs.  The  diagonal  cross  indicates  not  only 
that  the  rivet  head  is  to  be  countersunk,  but  that  it  is  also  to 
be  chipped  off  even  with  the  surrounding  material ;  if  the  rivet 
is  to  be  countersunk,  but  not  chipped,  the  countersunk  sign 
may  be  combined  with  the  sign  to  flatten  to  £  of  an  inch. 


42 


ARCHITECTURAL  ENGINEERING. 


Conventional   Signs   for   Rivets. 

Shop. 


Field. 


Two  full  heads . 


O 


Countersunk  inside  and  chipped. 


Countersunk  outside  and  chipped 


Countersunk  both  sides  and  chipped. . . 


Q 


Inside.         Outside. 


Both 
Sides. 


Flatten  to  ^  inch  high,  or  coun- 
tersunk and  not  chipped 


Flatten  to  |  inch  high 

Flatten  to  f  inch  high 


In  the  case  of  simple  flat  joints  the  front  side,  or  the  side 
which  is  seen,  is  considered  as  the  outside,  while  the  rear 
side,  or  the  side  which  is  hidden,  is  considered  as  the  inside. 

24.  Conventional  Signs  for  Rolled  Shapes. — When 
designating  the  rolled  structural  shapes,  such  as  angles, 
tees,  channels,  and  I  beams,  they  should  be  represented  by 
their  conventional  signs.  For  instance,  a  6"x6"x£"  angle, 
weighing  20  pounds  per  foot,  would  be  designated  upon  the 


§  G  ARCHITECTURAL  ENGINEERING.  43 

drawing  as  1-6"  X  G"  X  -£-"-20  JL,  or,  to  be  more  explicit,  it 
could  be  written  1-G"  X  G"  X  -\"  L-20  pounds  per  foot.  If  two 
angles,  or  a  pair  of  angles  of  this  size,  are  used,  they  could 
be  designated  as  2-G"  X  G"  X-£-"-203  L's.  It  is  not  usual, 
however,  to  give  the  weight  of  an  angle,  when  its  thickness 
is  given,  consequently,  a  G"xG"X|-"  angle,  weighing  20 
pounds  per  foot,  would  generally  be  expressed  on  a  drawing 
as  1-6"XG"X  £•"  L.  A  12-inch  channel,  weighing  40  pounds 
per  foot,  would  be  marked  on  the  drawing  1-12 "-40  #  C. 
Sometimes  the  length  of  the  rolled  section  is  marked 
on  the  drawing,  together  with  its  size  and  weight,  as 
2-6"xG'/Xi"XlO/0//L's. 

The  above  remarks,  together  with  a  careful  study  of  the 
working  drawings  given  in  Figs.  25  and  2G,  are  sufficient  to 
give  the  student  such  information  as  will  enable  him  to 
make  a  businesslike  shop  drawing  of  any  structure  or 
structural  member  made  up  of  rolled-steel  sections. 


STRENGTH  OF  KTVETS  AND  PINS. 

25.  Rivets  and  pins  are  the  elements  by  which  the  differ- 
ent sections  and  members  of  a  steel  structure  are  bound  or 
tied  together.  Pins  are  also  sometimes  used  in  the  better 
class  of  timber  construction,  in  which  case,  however, 
the  tension  members  are  usually  made  of  steel  or 
wrought  iron.  Such  a  pin  connection  was  shown  in 
Fig.  9. 

Where  plates  or  rolled  shapes  are  joined  by  either 
pins  or  rivets,  as  in  Fig.  27,  there  is  more  or  less  friction 
between  the  several  parts,  which  acts  to  prevent  them  from 
being  pulled  apart.  This  is  especially  true  when  rivets 
are  driven  close  against  the  plates  while  hot;  in  cooling, 
they  contract  between  the  heads  and  bind  the  plates  tightly 
together. 

The  friction  between  plates  bound  together  by  rivets  and 
pins,  however,  is  a  very  uncertain  quantity  and  should  not 
be  considered  in  calculating  the  strength  of  the  joint. 


44 


ARCHITECTURAL  ENGINEERING. 


•Sway  Brace. 


Riveted  Connection 


FIG.  27. 


26.     Methods  of  Failure. — Riveted  joints  may  fail  either 
by  the  shearing  of  the  rivet  or  the  crippling  of  the  plates 


ARCHITECTURAL  ENGINEERING. 


45 


which  they  connect.  When  a  rivet  shears,  the  tendency  is 
to  cut  straight  through  it  across  its  section,  as  shown  at  (a) 
and  (b),  Fig.  28. 

Where  there  are  only  two  plates  connected,  as  shown  at  (a), 
the  tendency  is  to  cut  the  rivet  on  the  single  plane  a  b.  A 
rivet  in  this  position  is  said  to  be  in  single  shear. 

At  (b),  the  tendency  is  to  cut  through  the  rivet  on  both  the 
planes  ab  and  cd\  under  these  conditions  the  rivet  is  in 
double  shear,  and  it  is  evident  that,  since  the  rivet  will  shear 
across  at  two  places,  it  will  be  twice  as  strong  as  where  the 
tendency  is  to  shear  through  only  one  section. 

When  the  diameter  of  the  rivet  is  large  in  proportion  to 
the  thickness  of  the  — j 

plate,  the  joint  or  con- 
nection is  liable  to  fail 
by  the  crippling  of  the 
plate  as  shown  in  Fig. 
29.  This  occurs  when 
the  resistance  of  the 
plate  to  crushing  or 
crimping  is  less  than  the  resistance  of  the  rivet  to  shear. 

27.  Bearing  Value  of  Hi  vets. — The  condition  of  the 
plates  in  the  joint  shown  at  (a),  in  Fig.  30,  is  called  ordinary 

bearing,\\\\\\G  the  plate 
m  connected  as  shown 
at  (b)  is  said  to  be  in 
li'eb  bearing.  This  dis- 
tinction is  important, 
because  the  bearing 
value  of  a  web-plate  is 
greater  than  that  of  an 
outside  plate,  the  value 
for  web  bearing  being 
about  \  greater  than 
for  ordinary  bearing. 
As  the  tensile  strength  of  iron  and  steel  used  in  the  manufac- 
ture of  rivets,  pins,  and  plates  for  structural  work  is  more 


46  ARCHITECTURAL  ENGINEERING.  §  6 

easily  determined  by  tests,  and,  therefore,  better  known  than 
either  its  compressive  or  shearing-  strength,  it  is  customary 
to  use  this  as  a  basis  from  which  to  calculate  the  bearing 
value  of  plates  and  the  shearing  strength  of  rivets  and 
pins. 

Good  practice  assumes  that  the  compressive  strength  of 
steel  or  high-test  iron  is  about  jf  of  its  tensile  strength ; 
that  is,  if  the  safe  tensile  strength  of  the  material  per  square 
inch  of  section  is  15,000  pounds,  the  safe  compressive 
strength  may  be  taken  as  |f  of  15,000  =  13,000  pounds. 

The  shearing  strength  is  considered  to  be  f  of  the  com- 
pressive strength.  For  example,  if,  as  above,  the  tensile 
strength  is  15,000  pounds  and  the  compressive  strength 
13,000  pounds,  the  shearing  strength  becomes  f  of  13,000 
=  10,833  pounds  per  square  inch  of  section. 

In  order  to  determine  the  bearing  value  of  the  plates 
around  a  rivet  hole,  it  is  necessary  to  consider  the  bearing 
area  of  the  rivet  on  the  plate.  This  is  always  assumed  to  be 
the  product  of  the  diameter  of  the  rivet  multiplied  by  the 
thickness  of  the  plate.  Owing,  however,  to  the  support 
which  the  material  around  the  hole  receives  from  the  rest  of 
the  plate,  it  will  stifely  sustain  a  pressure  greater  than  the 
compressive  strength  of  the  material  when  not  so  supported. 
The  safe  bearing  strength  of  the  rivet  on  the  plate,  for 
ordinary  bearing,  is,  therefore,  assumed  to  be  1^  times  its 
compressive  strength,  and  for  web  bearing  the  safe  bearing 
strength  is  assumed  as  double  the  compressive  strength  of 
the  material. 

In  deducting  the  rivet  holes,  to  ascertain  the  net  section  of 
a  riveted  plate,  the  diameter  of  the  hole  is  taken  as  ^  inch 
larger  than  the  diameter  of  the  rivet. 

EXAMPI.K  1. — Two  pieces  of  structural  steel  are  joined  by  rivets,  as 
shown  in  Fig.  31.  If  the  tensile  strength  of  the  steel  is  60,000  pounds 
per  square  inch,  and  a  factor  of  safety  of  4  is  used,  what  is  the  safe 
strength  of  this  joint  ? 

SOLUTION. — The  safe  tensile  strength  of  the  steel  is  60,000 -H  4 
=  15,000  pounds  per  square  inch.  The  width  of  the  pieces  connected 
is  2 >  inches,  from  which  is  to  be  deducted  1  inch  for  the  rivet  hole, 
leaving  a  net  width  of  1|  inches,  which,  multiplied  by  the  thickness  of 


6 


ARCHITECTURAL  ENGINEERING. 


the  plate,  gives  a  net  area  of  H  X  *  =  .5625  square  inch.  Then, 
.5625  X  15,000  =  8,437  pounds,  the  strength  of  the  plate. 

Now,  to  determine  whether  the  strength  of  the  rivets  is  equal  to  the 
net  section  of  the  plate :  Taking  the  compressive  value  of  the  plate  as 
jf  of  15,000  pounds,  or  18,000  pounds  per  square  inch,  and  the  rivets 
being  in  ordinary  bearing,  the  safe  bearing  strength  is  13,000  X  U 
=  19,500  pounds  per  square  inch  of  bearing  area.  The  bearing  area  is 
|Xf  =  -328  square  inch,  therefore,  the  safe  bearing  strength  for 
one  rivet  is  19,500  X  -328  =  6,396  pounds,  and  for  the  two  it  is  2  X  6,396 
=  12,792  pounds. 

The  next  point  to  consider  is  the  resistance  of  the  rivets  to  shear : 
The  shearing  strength  of  the  steel  is  £  of  13,000  =  10,833  pounds  per 
square  inch.  The  area  of  a  |-inch  rivet  is  .601  square  inch,  which, 


FIG.  31. 

multiplied  by  10,833,  gives  6,510  pounds,  the  shearing  strength  of  1 
rivet.  The  total  resistance  to  shear  of  the  rivets  in  the  joint  is,  there- 
fore, 6,510  X  2  =  13,020  pounds. 

The  safe  resistance  of  the  three  elements  entering  into  the  strength 
of  the  joint  is,  therefore,  as  follows :  Resistance  of  net  section  of  the 
plate  =  8,437  pounds;  bearing  value  of  the  plate  =  12,792  pounds; 
shearing  strength  of  the  two  rivets  =  13,020  pounds;  from  which  it  is 
easily  seen  that  the  strength  of  the  joint  is  that  of  the  net  section 
of  the  plate,  8,437  pounds.  Ans. 

Since  the  bearing  value  of  the  plate  and  the  shearing  strength  of 
the  rivets  are  considerably  in  excess  of  this  amount,  it  appears  that 
the  rivets  are  large  for  the  joint,  and  it  is  probable  that  f-inch  diam- 
eter rivets  would  give  better  results. 

EXAMPLE  2. — One  of  the  tension  members  in  a  structure  is  connected 
as  shown  in  Fig.  32.  The  tension  bars  are  made  of  structural  steel 
with  a  safe  tensile  strength  of  15,000  pounds  per  square  inch,  (a)  What 
is  the  bearing  value  of  the  bar  c  1  (/>)  What  is  the  bearing  value  of  the 
two  bars  a  ? 


48 


ARCHITECTURAL  ENGINEERING. 


SOLUTION. — (a)  The  safe  compressive  strength  of  the  material  is 
if  of  15,000  =  18,000  pounds.  Then  the  bearing  value  of  the  bar  c, 
which  may  be  considered  as  a  web,  is  2  X  13,000  =  26,000  pounds  per 
square  inch.  The  bearing  area  of  the  pin  in  the  bar  c  is  4  X  1  =  4 
square  inches;  therefore,  the  bearing  strength  of  the  bar  is  26,000  X  4 
=  104,000  pounds.  Ans. 

(b}  As  the  piece  a  is  in  ordinary  bearing,  its  bearing  value  is  13,000 


Diam.of  Pin  4-. 


FIG.  32. 

X  1|  =  19,500  pounds  per  square  inch.  The  bearing  area  of  the  two 
bars  is2x4xf  =  5  square  inches ;  their  combined  bearing  strength 
is,  therefore,  19,500  X  5  =  97,500  pounds.  Ans. 

EXAMPLE  3. — Determine  the  safe  strength  of  the  riveted  joint  shown 


t 

+       1 

"11^ 

^ 

*J                   \W         ^^         V^ 

1 

^  Oiam.  Rivet's, 
cf 

t 

Q     Q     Q 

<r> 

I 

FIG.  83. 

in  Fig.  33,  in  which  the  plates  and  rivets  each  have  a  safe  tensile 
strength  of  16,000  pounds  per  square  inch. 

SOLUTION. — The  safe  tensile  strength  of  the  material  being  16,000 
pounds,  the  safe  compressive  strength  is  {§  of  16,000  pounds,  or  13,867 
pounds,  and  the  shearing  strength  of  the  rivets  is  £  of  13,867,  or  11,556 
pounds  per  square  inch  of  section.  The  area  of  the  section  of  a  |-inch 
diameter  rivet  is  .601  square  inch ;  therefore,  the  total  shearing  strength 
of  the  three  rivets,  each  of  which  is  in  double  shear,  is  2  X  -601  X  11,556 
X3  =  41,670  pounds. 

The  two  outside  plates  are  in  ordinary  bearing,  and  their  bearing 
value  is  1|  X  13,866  =  20,800  pounds  per  square  inch.  There  are  three 


§G  ARCHITECTURAL  ENGINEERING.  49 

rivet  holes  in  each  plate,  each  with  a  bearing  area  of  jj  X  I •  =  .328 
square  inch;  the  total  bearing  strength  of  the  two  plates  is,  therefore, 
20,800  X  .328  X  3  X  2  =  40,934  pounds. 

The  bearing  value  of  the  central  or  web-bearing  plate  at  one  rivet 
hole  is  2  X  13,867  pounds  =  27,734  pounds  per  square  inch,  and  the 
bearing  area  is  £X|  =  -650  square  inch;  the  total  bearing  strength 
for  the  three  rivets  is,  therefore,  27,734  X  -656  X  3  =  54,580  pounds. 

The  safe  tensile  strength  of  the  central  plate  is  equal  to  its  net  sec- 
tion multiplied  by  16,000,  the  safe  unit  tensile  strength  of  the  material. 
The  net  width  of  the  plate  is  3  in.  —  I  in.  =  2  inches,  and  its  net  area, 
2xf  =  1.5  square  inches;  therefore,  the  safe  strength  is  1.5  X  16,000 
=  24,000  pounds.  The  strength  of  the  two  outside  plates,  calculated 
in  the  same  manner,  is  found  to  be  24,000  pounds  also;  therefore,  it  is 
evident  that,  since  the  strength  of  the  net  section  of  the  plate  is  much 
less  than  either  the  strength  of  the  rivets  or  the  bearing  value  of  the 
plates,  it  determines  the  strength  of  the  joint,  which  is,  therefore, 
24,000  pounds.  Ans. 

28.  Table  of  Bearing1  Values  of  Rivets. — In  order  to 
avoid  the  necessity  of  calculating  the  shearing  value  of  the 
rivets  and  the  bearing  value  of  the  riveted  plates  and  rolled 


[•  —  End  Distance.             ,^S<de  Distance. 

9    O    O    O   Q      O      Q-r-V      Q      O         99QQQQQOQQ 
Q    Q    Q    Q    Q       Q       O       Q-*-O       O          OOOOOOOOOO 

Side  Distance.  '-^ 
\  End  Distance.                                  -*IJ«  End  Distance. 

"    " 

^—  ' 

"      "       "       ^      "       " 

^_____  •            "       Portion  of  a  Plate  Girder. 

FIG. 


sections,  the  table  "Values  of  Rivets"  has  been  prepared. 
It  will  be  noticed  that  the  areas  and  shearing  values  for 
both  double  and  single  shear  are  given  for  rivets  from 


inch  to  |-  inch  in  diameter. 

1-25 


50  ARCHITECTURAL  ENGINEERING.  §  6 

The  least  distance  that  rivets  may  be  placed  from  the  end 
or  side  of  a  plate  is  given  in  the  third  and  fourth  vertical 
columns  from  the  left-hand  side  of  the  table.  By  referring 
to  Fig.  34,  the  terms  "end  distance"  and  "side  distance" 
used  in  the  table  will  be  readily  understood. 

The  rules  given  in  the  lower  right-hand  corner  of  the  table 
will  fix  these  distances  and  should  be  borne  in  mind,  as  it 
may  not  always  be  convenient  to  refer  to  the  table. 

In  the  lower  left-hand  part  of  the  table  is  given  all  the 
necessary  rules  for  determining  the  several  values  as  tab- 
ulated. 

The  values  under  ' '  allowable  stress  per  square  inch  for 
high-test  iron  or  structural  steel,"  refer  to  the  tensile 
strength  of  the  material.  If  an  allowable  stress  of  15,000 
pounds  is  desired,  use  the  value  of  the  plates  and  rivets  given 
under  this  head ;  should  a  still  safer  connection  be  desired, 
and  an  allowable  stress  of  10,000  pounds  per  square  inch  be 
considered  advisable,  use  the  values  given  in  the  column 
headed  10,000. 

If  the  requirements  of  the  structure  are  such  that  values 
other  than  those  given  in  the  table  are  desired,  the  required 
values  may  be  obtained  by  finding  the  sum  or  the  difference 
of  two  of  the  given  columns.  For  example,  a  stress  of  16,000 
pounds  per  square  inch  is  to  be  allowed,  and  the  ordinary 
bearing  value  of  a  ^-inch  plate  at  a  -f-inch  rivet  hole  is 
required ;  by  adding  the  value  given  for  an  allowable  stress 
of  1,000  pounds  to  that  given  for  15,000  pounds,  the  required 
bearing  value  may  be  determined ;  in  this  case  it  is  203. 13 
+  3,046  =  3,249.13,  which  is  the  allowable  bearing  value  of 
a  ^-inch  plate  in  ordinary  bearing  around  a  f-inch  diameter 
rivet,  where  the  safe  tensile  stress  of  the  materials  is  taken 
at  16,000  pounds  per  square  inch. 

29.  Pins  Subjected  to  Bending  Stresses. — Pins,  when 
used  to  connect  the  several  •  members  of  a  structure,  besides 
being  subjected  to  shearing  in  the  same  manner  as  rivets, 
may  be  required  to  resist  heavy  bending  stresses ;  they  may 
then  be  regarded  as  solid  cylindrical  beams  and  calculated 


6 


ARCHITECTURAL  ENGINEERING. 


51 


to  resist  the  greatest  bending  moment  that  may  come  upon 
them. 

Take,  for  example,  the  pin  in  Fig.  35  which  connects  the 
three  tension  bars  «,  «,  and  c\  the  pull  of  the  two  bars  a,  a, 


SOOOOlb. 


Backing  P/afes  or  Washers. 


~\ 


30OOOlb. 


FIG.  .35. 


both  acting  in  the  same  direction,  is  transmitted  to  the  bar 
c  by  means  of  the  pin.  The  stress  upon  each  bar  a  is  30,000 
pounds,  consequently  the  stress  upon  the  bar  c  must  be 
60,000  pounds.  Assuming  a  maximum  unit  stress  of  15,000 
pounds  per  square  inch,  it  is  desired  to  find  what  diameter 
of  pin  is  required  to  resist  the  bending  moment  produced  by 
the  stresses  exerted  upon  it. 

In  calculating  the  bending  moment  on  a  pin,  the  forces 
acting  upon  it  through  the  several  members  are  considered 
as  being  applied  at  the  center  of  the  bearings.  In  Fig.  35, 
the  distance  between  the  centers  of  the  bearings  of  the  mem- 
bers is  4  inches,  and  by  referring  to  the  diagram,  Fig.  30, 

30000  lb». 


30000  Ibs. 


6000O  Ibs. 


FIG.  36. 


it  is  seen  that  the  greatest  bending  moment  is  at  c,  and  is 
equal  to  30,000X4  —  120,000  inch-pounds. 

Having-  found  the  greatest  bending  moment  on  the  pin, 
it  is  necessary  to  determine  its  diameter,  in  order  that  its 
resisting  moment  may  equal  the  moment  of  the  bending 
stresses. 


52  ARCHITECTURAL  ENGINEERING.  §  6 

In  the  table  "Elements  of  Usual  Sections"  we  find  the 
section  modulus  of  a  circular  section  to  be 


K  = 


AD       .7854£>*x£> 


=  .0982Z>3; 


8  8 

the  allowable  unit  stress  on  the  material  is  S  =  15,000 
pounds  per  square  inch,  and  the  bending  moment  is  M 
—  120,000  inch-pounds;'  substituting  these  values  in  formula 
4,  Art.  1O,  we  have  M  -  SK,  or  120,000  =  15,000 
X-0982  /}3,  from  which  we  get 

120,000         = 
15, 000  X. 0982 

The    diameter    of    the    pin     is,    therefore,    D  =  ^8 1.46 
=  4.335  =  4|  inches,  nearly. 

30.  Table  of  Resisting  Moments  of  Pins. — To  avoid 
the  necessity  of  calculating  the  resisting  moment  of  pins,  the 
table  "  Resisting  Moments  of  Pins  "  will  be  found  convenient. 
This  table  gives  the  resisting  moments  of  pins  from  1  to  12 
inches  in  diameter,  calculated  for  allowable  fiber  stresses  of 
15,000,  20,000,  22,000,  and  25,000  pounds  per  square  inch. 

31.  Resultant  Moment  of  Several  Stresses. — When 


a  pin  is  used  at  a  joint  at  which  several  members,  extending 


6 


ARCHITECTURAL  ENGINEERING. 


in  different  directions,  meet,  as  shown  in  Fig-.  37,  it  is  neces- 
sary to  combine  the  stresses  so  as  to  find  the  resultant  that 
gives  the  greatest  bending  moment.  This  is  conveniently 
done  by  first  resolving  the  stresses  on  each  of  the  different 
members  into  vertical  and  horizontal  components,  and  cal- 
culating the  bending  moments  produced  in  each  of  these 


2OOOO  Ib. 


Compressive  Stress 
in  this  Member  4OOOO  Ib. 


Tensile  Stress 
t'n  these  Members  6OOOO  /{>. 


I  Jiooj 

T^j  _  ^jj"!^  — 

rp     ? 

IM~  T  " 

r                               oa 

-jrn^TJ  aJHUi'  

isi 

ny  Pieces  or  Washers  /eft  out. 


Fin.  38. 


directions  by  all  the  corresponding  components.  The  max- 
imum bending  moment  is  then  given  by  the  resultant  of 
these  two  bending  moments. 

The  details  of  the  method  for  finding  the  maximum  bend- 
ing stress  on  a  pin,  will  be  made  clear  by  a  study  of  the  fol- 
lowing illustrative  examples: 

Fig.  38  shows  one  of  the  lower  joints  of  a  roof  truss.     At 


54 


ARCHITECTURAL  ENGINEERING. 


6 


this  joint  there  are  four  sets  of  members,  two  of  which  act 
in  a  horizontal,  while  one  acts  in  a  vertical  direction;  since 
they  already  act  in  the  directions  of  the  required  compo- 
nents, these  forces  need  not  be  resolved.  There  is,  how- 
ever, one  inclined  member  in  which  there  is  a  compressive 
stress  of  40,000  pounds,  which  stress  must  be  resolved  into 
its  vertical  and  horizontal  components.  Draw  the  line  a  b 
parallel  to  the  strut  and  of  such  a  length  as  to  represent  the 
magnitude  of  the  stress.  From  «,  draw  the  horizontal  line 
a  c  intersecting  the  vertical  line  at  the  point  c.  The  direc- 
tion of  the  forces  around  the  triangle  is  shown  by  the 
arrows.  Upon  measuring  the  line  a  c,  the  horizontal  com- 
ponent of  the  stress  in  ab  is  found  to  be  20,000  pounds, 
while  the  vertical  component  of  the  stress  is  found  to  be 
34,650  pounds. 

Having  determined  these  components,  a  diagram  showing 
all  the  horizontal  stresses  acting  upon  the  pin  and  tending 
to  bend  it,  and  also  another  showing  all  the  vertical  forces, 

should  be  drawn  as  il- 
lustrated at  (a)  and  (b), 
Fig.  39,  the  distance 
from  center  to  center 
of  the  members  being 
taken  from  the  detail 
plan  of  the  joint,  Fig. 
38.  It  must  always  be 
remembered  that,  in 
accordance  with  the 
principles  of  equilib- 
rium, the  sum  of  the 
resultants  of  the  forces 
acting  upon  the  pin  in 
any  one  direction  must 
equal  the  sum  of  all 
the  resultants  acting  in 

the  opposite  direction; 
' 
otherwise,   the    pin 

would  move  in  the  direction  of  the  greater  sum,  and  the 


3OOOO  Ib. 


cl 

1 

s 

30OOO  Ib. 

H 

1OOOO  Ib.    y 

2OOOO  Ib. 

f. 
' 

20OOO  Ib.      ^ 

I1 

1OOOO  Ib.    ^ 

(a) 

17325  Ib. 

17325  Ib. 


34650  Ib. 


(b) 


FIG.  39. 


§  o  ARCHITECTURAL  ENGINEERING.  55 

structure  would  fall.  Thus,  from  Fig.  38,  it  is  readily 
seen  that  the  vertical  component  of  the  stress  in  />'  C  acts  in 
an  opposite  direction  to  the  stress  in  the  member  CD, 
while  the  horizontal  component  acts  in  opposition  to  the 
stresses  in  the  member  A  B,  and  in  the  same  direction 
as  the  stress  in  D  E.  This  makes  the  algebraic  sum  of 
all  the  components  in  either  the  horizontal  or  vertical 
direction  equal  to  zero,  and  fulfils  the  condition  of 
equilibrium. 

The  resultant  of  the  vertical  and  horizontal  bending 
moments  may  also  be  calculated  by  the  rule  for  finding  the 
length  of  the  hypotenuse  of  a  right-angled  triangle;  for 
example,  in  this  case  the  lengths  of  the  sides  are  represented 
by  the  horizontal  bending  moment  of  50,000  inch-pounds, 
and  the  vertical  bending  moment  of  69,300  inch-pounds;  the 
resultant  bending  moment  is,  therefore,  V/50,OUU2  +  69,300'' 
—  85,454  inch-pounds. 

In  order  to  determine  the  required  size  of  pin  for  this 
joint,  assume  a  safe  fiber  stress  of  15,000  pounds  per  square 
inch,  then,  by  referring  to  the  table  "Resisting  Moments 
of  Pins,"  it  is  seen  that  a  pin  3|  inches  in  diameter,  under  a 
fiber  stress  of  15,000  pounds  per  square  inch,  has  a  resisting 
moment  of  85,700  inch-pounds,  which  is  very  nearly  the 
value  required  by  the  conditions. 

The  student  must  remember  that  in  all  cases  the  pin 
should  be  examined  for  both  shear  and  bending  stresses. 


EXAMPLES  FOR  PRACTICE. 

1.  What  are  the  safe  strengths  of  |,  |,  and  f  inch  rivets,  in  double 
shear,  and  also  in  single  shear,  assuming  that  the  safe  tensile  strength 
of  the  material  used  in  their  manufacture  is  15,000  pounds  per  square 
inch  of  section  ? 

Diameter  of  Rivet.        Double  Shear.  Single  Shear. 

Inch.                           Pounds.  Pounds. 

I                           13,022  6,511 

Ans.  -I                 »                           9,577  4,788 

6,652  3,326 


ARCHITECTURAL  ENGINEERING. 


m  Rivets. 


2.  What  pulling  force  will  two  pieces  of  f "  X  2  J"  bar  safely  resist,  pro- 
viding they  are  connected 
at  the  ends  by  two  |-inch 
diameter  rivets,  as  shown  in 
~S  Fig.  40?  The  safe  tensile 
strength  of  the  material  in 
rivet  and  bar  is  15,000 
pounds. 

*£Wrt.  Iron  Bar.  Ans.    9,140  lb. 

3.  What  is  the  safe  resist- 
ing moment  of  a  pin  5  inches  in  diameter,  if  the  safe  fiber  strength  of 
the  material  is  20,000  pounds  ? 

Ans.  245,400. 

4.  In  Fig.  41  is  shown  a  pin  connection,  the  pull  on  the  tension  bar 
a  being  140,000  pounds.  If  the  safe  shearing  strength  of  the  material 
in  the  pin  is  10,000  pounds  per  square  inch,  and  the  safe  fiber  stress  in 


FIG.  41. 

bending  is  15,000  pounds  per  square  inch,  (a)  what  size  of  pin  will  be 
required  to  resist  trie  shear  ?  (b)  what  size  will  be  required  to  resist  the 
bending  ? 

A          (  (a)    3  in.  in  diameter. 
(  (b)    4£  in.  in  diameter. 

5.     It  is  necessary  to  construct  the  connection  of  a  tension  member 
as  shown  in  Fig.  42.     What  is  the  safe  load  that  this  member  will 


FIG.  42. 


Wrt  Iron  Bar. 


carry,  if  the  safe  tensile  strength  of  the  material  in  both  the  rivets  and 
bars  is  18,000  pounds  per  square  inch  ? 

Ans.  ll,2501b. 


ARCHITECTURAL  ENGINEERING. 


57 


PL.AT.K 


GTCNKRA  TJ    CONSTRUCTION. 

32.  A  plate  girder  is  a  beam  built  up  of  a  number  of 
plates  and  angles  securely  riveted  together. 

The  names  given  to  the  different  parts  of  a  plate  girder 
may  be  understood  by  referring  to  Fig.  43, 
in  which  a  is  the  flange  plate,  of  which 
there  may  be  one  or  more  on  each  flange, 
depending  upon  the  strength  required. 
The  flange  plates  are  the  principal  ele- 
ments for  resisting  the  bending  stresses  in 
the  girder.  The  flange  angles  /;,  /;  are 
the  means  of  connecting  the  flange  plates 
and  the  web-plate  c.  When  the  load  on 
the  girder  is  small,  the  flange  plates  may 
be  omitted,  in  which  case  the  flange  angles 
are  the  members  which  chiefly  act  to  resist 
the  bending  stresses. 

33.  Stiffness. — On  account  of  the  con-  pio. -43. 
struction  of  a  plate  girder,  there  is  very  little  stiffness  in  the 
web-plate,  consequently,  there  is  always  a  strong  tendency 
for    it    to   fail    by    buckling    and    twisting    under    the    load 
imposed    upon    the    girder.       This    tendency    to   buckle    is 
greatest  at  the  supports  or  abutments  of  the  girder  and  at 
points  where  concentrated  loads  are  applied.     Because  of 
this  buckling  tendency  it  becomes  necessary  to  reinforce  the 
girder  by  riveting  at  stated  intervals  stiffeiiers  generally 
made  of  angles. 

The  most  common  and  cheapest  form  of  stiffener  is  shown 
at  (a),  Fig.  44.  This  is  simply  a  straight  piece  of  angle 
riveted  to  the  web-plate  and  flange  angles.  The  space 
between  the  stiffeners  and  web-plate,  due  to  the  thickness 
of  the  flange  angles,  is  filled  with  apiece  of  bar  iron  or  plate, 
as  shown  at  d\  this  is  called  a  filler  or  packing:  piece. 


58 


ARCHITECTURAL  ENGINEERING. 


6 


A  more  expensive  form  of  stiffener  for  plate  girders  is 
shown  at  (£),  Fig.  44.  The  angle  is  swaged  out,  to  allow 

it  to  fit  over  the  flange  an- 
gles, and  is  riveted  directly 
to  the  web-plate,  thus  do- 
ing away  with  the  filler  or 
packing  piece.  This  con- 
struction does  not  require 
as  much  material  as  that 
shown  at  (a),  but,  unless 
there  are  a  large  number 
of  girders  of  the  same 
dimensions  to  be  built,  in 
which  case  dies,  in  connec- 
tion with  a  power  or  hy- 
draulic press,  may  be  used 
for  swaging  the  ends,  the 
labor  required  is  so  much 
greater  as  to  make  the 
girder  more  expensive. 

The  stiffeners  shown  at 
(c)  are  sometimes  used, 
but  are  subject  to  the 
same  general  criticism  in  regard  to  cost  of  manufacture  as 
those  shown  at  (b).  Stiffeners  of  this  shape  possess  a  pos- 
sible advantage  in  the  fact  that  they  stiffen  the  flanges 
considerably  more  than  either  of  the  other  two  styles. 

34.  Usual  Forms  of  Sections. — The  four  principal 
sections  used  in  plate-girder  construction  are  shown  in 
Fig.  45. 

A  simple  plate  girder  with  a  web-plate  and  two  flange 
angles,  but  with  no  flange  plate,  is  shown  at  (a).  This  sec- 
tion is  used  for  short  spans  or  light  loads.  At  (b)  is  shown 
a  similar  girder  with  one  flange  plate.  This  girder  is  used 
to  support  heavier  loads  and  to  clear  longer  spans;  while 
the  girder  at  (c),  which  may  have  two  or  more  flange  plates 
at  each  flange,  may,  if  the  conditions  require,  be  made  as 


FIG.  44. 


6 


ARCHITECTURAL  ENGINEERING. 


50 


heavy  as  is  necessary  in  order  to  carry  great  loads  over 
long  spans.  In  fact,  the  strength  of  a  girder  of  this  char- 
acter may  be  increased  almost  indefinitely  by  adding  flange 


(a) 


plates.  The  section  (^/),  Fig.  45,  is  a  plate  girder  of  box 
section.  It  is  stiffer  laterally  than  the  forms  shown  at  (a), 
(&),  and  (c),  but  the  difficulty  of  reaching  the  interior  for 
painting  and  inspecting,  and  the  excessive  amount  of  labor 
required  in  its  construction,  are  such  serious  objections  that 
it  is  much  less  used  than  the  sections  shown  at  (a),  (£),  and 
(c).  On  account  of  their  open  construction,  the  latter  are 
especially  good  forms  to  use  in  buildings  where  the  objec- 
tion in  regard  to  lateral  stiffness  does  not  hold  good,  as 
when  the  girder  is  used  in  the  position  in  which  it  is  usually 
found, -being  generally  prevented  from  deflecting  laterally 
by  the  floorbeams;  any  lack  of  stiffness  in  comparison  with 
the  box  girder  is  more  than  compensated  by  the  simplicity 
of  construction  and  easy  access  on  all  sides  for  painting  and 
inspecting. 


PRINCIPLES    OF    DESIGN. 


STRESSES. 

35.  The  external  forces,  loads,  and  reactions  produce 
the  same  kind  of  stresses  in  a  plate  girder  as  in  an  ordinary 
beam,  but,  on  account  of  its  special  construction,  the  distri- 
bution of  these  stresses  in  the  girder  is  assumed  to  be  some- 
what different  from  that  in  a  beam  made  of  a  single  piece. 


GO  ARCHITECTURAL  ENGINEERING.  §  6 

In  the  girder,  the  shear  is  generally  assumed  to  be  borne 
wholly  by  the  web-plate,  while  the  bending  moment  is 
assumed  to  be  resisted  by  the  stresses  in  the  flange  mem- 
bers. The  method  of  calculating  the  magnitude  of  the  shear 
and  bending  moment  is  the  same  as  that  for  beams,  already 
discussed  in  Architectural  Engineering,  §  5 ;  owing,  however, 
to  the  different  assumption  in  regard  to  the  distribution  of 
these  forces,  a  different  method  of  calculation  is  used  in 
determining  the  relations  between  them  and  the  stresses  in 
the  girder. 

36.  Shearing  Stresses  in  Web-Plate. — In  discussing 
the  methods  of  calculating  the  dimensions  of  a  plate  girder 
for  a  given  purpose,  we  will  first  consider  the  shear,  which 
is  the  principal  factor  that  determines  the  thickness  of  the 
web-plate  and  the  number  and  size  of  stiffeners  required. 
In  Architectural  Engineering,  §5,  it  was  shown  that  the 
greatest  shear  in  a  beam  occurs  at  the  point  of  support  at 
which  the  reaction  is  greatest,  and  that  the  magnitude  of 
the  shear  is  equal  to  the  reaction  at  that  point ;  consequently, 
in  a  simple  plate  girder,  the  greatest  shear  occurs  at  a  point  of 
support,  and  is  equal  in  amount  to  the  reaction  at  that  point. 


WEB-PLATES  AND  STTFFENEKS.  • 

37.  Depth  of  Girder. — Having  calculated  the  shear, 
the  depth  of  the  girder  is  assumed  in  accordance  with  prac- 
tical rules  which  fix  the  relation  between  the  depth  and 
span.  In  accordance  with  the  best  practice,  the  depth 
should  not  be  less  than  y1^-  of  the  span,  though  some  author- 
ities consider  -^  as  being  ample.  The  latter  proportion, 
however,  gives  an  exceedingly  shallow  girder,  and  cannot  be 
recommended  except  where  the  loads  are  very  light  and  the 
span  short,  or  where  it  is  absolutely  necessary  that  an 
extremely  shallow  girder  be  used,  on  account  of  decorative 
features,  or  lack  of  space  in  regard  to  headroom,  in  which 
case  the  girder  should  be  so  proportioned  that  when  fully 
loaded  its  deflection  will  not  be  excessive. 


A  RC H I T ECT U  R A L  EXG I X  E E R I XG. 


38.  Thickness  of  Web-Plate. — Knowing  the  depth 
of  the  girder,  and  the  shear  at  the  points  of  support, 
the  thickness  of  the  web-plate  is  proportioned  so  as  to 
give  it  sufficient  area  to  resist  the  maximum  shear.  It 
is  always  necessary  to  stiffen  the  plates  over  the  supports, 
as  shown  in  Fig.  47;  these  stiffeners  are  riveted  to  the 
plate  and  transfer  the  shearing  stress  from  it  to  the 
supports. 

A  considerable  portion  of  the  plate  is  cut  away  by  the 
holes  for  the  rivets  bv 

n 

which  it  is  fastened  to  the 
stiffeners;  hence,  the  least 
strength  of  the  plate  is 
along  the  line  of  the  rivet 
holes.  It  can  readily  be  \ 
seen,  by  referring  to  Fig. 
46,  which  shows  the  end 
of  a  plate  with  the  holes 
punched  for  riveting  to  the 
stiffener,  that  the  net  or 
efficient  depth  of  the  plate 
is  equal  to  the  actual  depth 
minus  the  sum  of  the  diam- 
eters of  the  rivet  holes. 

The  following  rule  may 
be  used  for  calculating  the 
thickness  of  a  web-plate  so 
that  it  will  have  sufficient 
strength  to  resist  the  shear- 
ing stress : 

Rule. — From  the  total  depth  of  the  web-plate,  deduct  tlie 
sum  of  the  diameters  of  the  rivet  holes,  which  id  I  I  give 
the  net  or  efficient  depth  of  the  web-plate;  multiply  the  net 
depth  by  the  safe  resistance  of  the  material  to  shear,  and  divide 
the  maximum  shear  in  pounds  by  the  product;  the  quotient 
will  be  the  required  thickness  of  the  metal  in  the  web  of  the 
girder. 


s/  i-ja//am  Rivet  Ho/es. 

i                    ; 

, 

y 

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o 

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I 

i 
it 

; 

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(0 

; 

rv 

i              f 

; 

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c 

) 

t 

( 

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i 

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( 

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\ 

! 

j 

( 

; 

5 

f 

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c 

r 

Secf/'on  on  (t  b 

ft 

Fir..  -16. 

62 


ARCHITECTURAL  ENGINEERING. 


The  above  rule  may  be  expressed  by  the  formula 

r> 


in  which  T  =  thickness  of  the  web-plate ; 

R  =  greatest  reaction  or  maximum  shear; 

5  =  safe  shearing  resistance  of  the  material  per 

square  inch; 
D  =  net  depth  of  the  web-plate  after  all  the  rivet 

holes  have  been  deducted. 

The  safe  resistance  of  the  material  to  shear  is  of  course 
governed  by  the  factor  of  safety  required  in  the  girder.  For 
example,  the  ultimate  shearing  strength  of  structural  steel 
being  52,000  pounds  per  square  inch,  if  a  factor  of  safety  of 
4  is  required,  the  safe  resistance  of  the  metal  will  be  52,000 
-=-4  =  13,000  pounds,  while  if  a  factor  of  safety  of  5  is 
desired,  the  safe  strength  will  be  52,000  -f-  5  =  10,400  pounds. 

In  deducting  the  metal 
for  the  rivet  holes  in  order 
I    to  ascertain  the  net  depth 
of  the  web-plate,  the  holes 
\  should  always  be   consid- 
)  ered  as  being  \  inch  larger 
f  than  the  nominal  diameter 
of  the    rivet;    this    allow- 
ance is  made  because  the 
holes  are  always  made  -^ 
inch    larger    in    diameter 
than  the  rivet  so  that  the 
rivet  may  be  inserted  eas- 
ily,   and   another  -fa  inch 
should  also  be  allowed  in 
the  diameter  of  the  hole  to 
compensate  for  any  injury 
that    the    metal    immedi- 
ately around  it  may  suffer 
FIG.  47.  f          .  •,  i 

from  the  punch. 

It  will  often  be  found  that  the  calculated  thickness  of 
the  web-plate  is  less  than  is  allowable  for  practical  reasons. 


^ 

Q       Or 

•G- 

o      o      o 

J 

O 

/ 

J 

O 

( 

J 

O 

*     \ 

(3 

J 

o 

1 

i 

o 

« 

'J 

<j 

o 

>      / 

a 

0 

£ 

j 

•• 

o 

1" 

j 
j 

Q 

o 

<o 

j 

Q 

j 

O      O*- 

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o     o      o          / 

•V\-«  ';§'••  4£>y^  Al1  Rivets^diam. 

I 

I         l 

-|  1—       -T           f    '  

Reaction  at  this  Support  i £0000  Ib. 


§  G  ARCHITECTURAL  ENGINEERING.  <i3 

The  thinnest  plate  that  should  be  used  for  any  case  is  T\ 
inch. 

EXAMPLE. — Fig.  47  shows  the  end  of  a  plate  girder  in  which  the 
greatest  reaction  is  120,000  pounds.  The  girder  is  made  of  structural 
steel,  the  safe  fiber  stress  of  which  is  assumed  to  be  11,000  pounds  per 
square  inch  for  shear.  What  should  be  the  thickness  of  the  web-plate  ? 

SOLUTION. — The  width  of  the  plate  is  48  inches,  and  there  are  13  holes 
punched  for  |-inch  rivets.  The  deduction  to  be  made  for  each  rivet 
hole  is  i  inch  +  f  inch  =  f  inch,  therefore,  the  net  depth  of  the  plate 
is  48  —  13  X  |  =  48  —  llf  =  36 1  inches.  Applying  formula  12,  the 
thickness  of  the  plate  is 

120,000 
r=86|X  11,000  =-29'mdl- 

In  no  case,  however,  should  the  web-plate  of  a  girder  be  less  than 
•fg  inch  in  thickness.  Hence,  as  .297  is  less  than  yV,  the  thickness  of 

16  1  t> ' 

the  web-plate  in  this  girder  should  be  -fa  inch.     Ans. 

39.  Buckling  of  Web-Plate  and  Distribution  of 
Stiffeiiers. — The  shearing  stresses  in  a  web-plate,  in  addi- 
tion to  their  tendency  to  shear  the  plate,  as  above  described, 
are  liable  to  cause  it  to  fail  by  buckling;  therefore,  in  order 
to  properly  resist  the  vertical  shearing  stresses  and  prevent 
them  from  buckling  the  web-plate  before  its  full  shearing 
strength  is  realized,  it  is  necessary  to  provide  the  stiffeners, 
which  have  been  previously  described. 

It  was  shown  in  Architectural  Engineering,  §  5,  that  the 
shearing  stresses  in  a  simple  beam  are  always  greatest  at  the 
points  of  support,  and  diminish  towards  the  center  of  the 
beam  tmtil  a  point  is  reached  between  which  and  the  sup- 
port the  sum  of  the  loads  is  equal  to  the  reaction;  at  such  a 
point  the  shear  is  said  to  change  sign.  The  natural  infer- 
ence from  the  above  fact  is  that  the  stiffeners  should  be 
more  numerous  at  the  points  of  greatest  vertical  shear, 
decreasing  in  number  as  the  shear  decreases.  Theoretically, 
this  would  be  a  correct  method  of  locating  the  stiffeners, 
but  practically  they  are  spaced  at  equal  distances  along  the 
length  of  the  girder,  except  at  the  points  of  support,  where 
several  are  placed  near  each  other  in  order  to  give  the  end  of 
the  girder  more  nearly  the  character  of  a  column  and  enable 


64  ARCHITECTURAL  ENGINEERING.  §  6 

it  to  successfully  resist  the  great  vertical  shear,  due  to  the 
reaction  at  this  point.  In  no  case  should  the  stiffeners  at 
the  end  of  a  plate  girder  be  omitted,  even  if  the  conditions 
make  the  intermediate  ones  unnecessary. 

It  is  also  good  practice  to  place  stiffeners  directly  under 
any  concentrated  load  that  may  be  placed  upon  the  girder. 
These  stiffeners  are  necessary  not  only  to  stiffen  the  web- 
plate  at  the  point  of  application  of  the  load,  and  thus  prevent 
buckling,  but  also,  through  the  medium  of  the  rivets,  to 
assist  in  distributing  the  load  on  the  web-plate  and  other 
members  of  the  girder. 

The  end  stiffeners  of  a  plate  girder  may  be  considered  as 
columns  subjected  to  a  compressive  stress  equal  to  the  reac- 
tion, and  calculated  by  the  rules  and  formulas  already  given 
for  columns.  For  safety  the  stress  upon  the  end  stiffeners 
should  never  exceed  15,  000  pounds  per  square  inch  of  section. 

Practice  in  regard  to  the  placing  of  stiffeners  on  plate 
girders  varies  considerably,  being-  more  a  matter  of  judg- 
ment and  experience  than  of  calculation.  Some  engineers 
determine  the  resistance  of  the  web-plate  to  buckling  by  the 
formula 

.     _        11,000  (13.) 

= 


3,000  *a 

in  which  B  —  safe  resistance  of  the  web  to  the  buckling,  in 

pounds  per  square  inch  ; 
d  —  depth  of  the  web-plate  in  inches; 
t  =  thickness  of  the  web-plate  in  inches. 
If  the  value  of  B  given  by  this  formula  is  less  than  the 
unit  shearing  stress,  the  girder  should  be  stiffened. 

EXAMPLE.  —  The  allowable  shearing  stress  upon  the  web  of  a  plate 
girder  is  11,000  pounds  per  square  inch.  The  stiffeners  at  the  end 
supports  are  riveted  by  nine  |-inch  rivets  to  the  web-plate,  which  is 
36  inches  wide.  The  end  reaction  on  the  girder  is  100,000  pounds. 
(a)  What  should  be  the  thickness  of  the  web-plate  ?  (b)  Will  it  be  suffi- 
ciently strong  without  the  addition  of  stiffeners  ? 

SOLUTION.  —  (a)  Since  |-inch  rivets  are  used,  the  allowance  to  be 
made  for  one  rivet  hole  is  |  +  £  =  1  inch,  and  the  effective  width  of  the 


§  0  ARCHITECTURAL  ENGINEERING.  Go 

plate  along  the  line  of  rivets   is  36  —  9  X  1  =  27  inches.      Applying 
formula  lli,  the  thickness  of  the  plate  is  found  to  be 

100,000 
T=  27  X  11.000 

The  thickness  of  the  nearest  standard-size  plate  above  this  is  j|  inch, 
which  will  be  the  thickness  used.     Ans. 

(b)  By  formula  13,  the  safe  unit  resistance  of  the  plate  to  buckling  is 
11,000 


=  2,100  pounds  per  square  inch; 


1 


3,000  X  if 


which,  since  it  is  much  less  than  the  unit  shearing  stress  on  the  web- 
plate,  shows  that  stiffeners  are  required.  Ans. 

4O.     Practical  Rule  for  Spacing  StifFeiiers.  —  It  is  not 

the  general  practice  to  make  the  above  calculations  to  deter- 
mine whether  stiffeners  are  required  ;  according  to  the  best 
engineering  practice,  stiffeners  should  be  provided,  unless 
the  thickness  of  the  web-plate  is  at  least  Jj-  of  the  clear  dis- 
tance between  the  vertical  legs  of  the  flange  angles. 

EXAMPI.K.  —  Assume  the  girder  in  the  previous  problem  to  be  pro- 
vided with  6"  X  6"  flange  angles;  the  depth  and  thickness  of  the  plate, 
as  previously  shown,  are  36  inches  and  £  inch,  respectively.  According 
to  the  above  rule,  does  this  girder  require  stiffeners  ? 

SOLUTION.  —  The  unsupported  depth  of  the  plate  between  the  flange 
angles  is  36  —  2  X  6  =  24  inches  ;  -s\j  of  24  inches  =  .48,  say,  \  of  an  inch. 
As  the  thickness  of  the  web-plate  is  only  f  of  an  inch,  the  girder  must 
be  provided  with  stiffeners.  Ans. 

Another  rule,  which  gives  nearly  the  same  result,  is  as 
follows: 

Rule.  —  Provide  stiffeners  whenever  the  thickness  of  the 
web-plate  is  less  than  -^  of  its  total  deptJi. 

This  rule  is  modified  by  some  authorities  so  as  to  allow 
a  thickness  of  -^  of  the  total  depth  of  the  plate  as  being 
amply  safe  without  stiffeners.  The  more  conservative  rule, 
however,  which  requires  the  thickness  to  be  at  least  -£•$  of 
the  unsupported  depth  of  the  web  or  the  distance  between 
the  flange  .angles,  is  the  one  to  be  recommended,  and  will 
be  used  in  this  section. 

The  spacing  and  size  of  stiffeners  to  be  used  on  a  plate 

1-26 


66 


ARCHITECTURAL  ENGINEERING. 


6 


Secf/on  on   a  b 


girder  is  almost  entirely  a  matter  of  experience  and  judg- 
ment. As  a  general  rule,  it  may  be  said  that  stiffeners 
should  be  provided  at  the  ends  of  all  plate  girders  over  the 
supports  or  abutments,  and  they  should  be  so  proportioned 
that  they  will  take  care  of  the  entire  reaction  at  these  points. 
The  stiffeners  between  the  abutments  or  supports  should  be 
of  such  a  size  that  they  will  best  suit  the  general  require- 
ments of  the  design  of  the  girder.  The  practice  in  spacing 
intermediate  stiffeners  is  to  make  the  distance  between  their 
center  lines  equal  to  the  depth  of  the  girder,  thus  dividing 

the  girder  into  equal 
square  panels.  Under 
no  conditions,  however, 
should  stiffeners  be 
placed  more  than  5  feet 
apart  from  center  to 
center  of  line  of  rivets. 

Having  proportioned 
the  stiffeners  at  the 
abutments  to  take  the 
entire  reaction,  it  is 
good  practice,  when 
possible,  to  make  the 
intermediate  stiffeners 
of  the  same  size  as  the 
end  ones.  In  general, 
the  angles  used  for  stif- 
feners should  not  be  less 
-  than  3  in.  X  3  in.  X  & 
in.,  though  on  shallow 
girders,  with  extremely 
light  loads,  it  might  be 
economical  to  use  an- 
gles as  light  as  2|-  in. 
FIG.  48.  x  2£  in.  X  ^  m-  Sizes 

smaller  than  this  should  certainly  never  be  used  as  stiffeners. 
Stiffeners  should  always  extend  over  the  vertical  legs  of 
the  flange  angles ;  they  should  always  be  either  swaged  out 


§  6  ARCHITECTURAL  ENGINEERING.  «7 

to  fit  over  the  flange  angles,  or  be  provided  with  a  filling- 
piece  as  illustrated  in  Fig.  44. 

EXAMPLE. — The  end  reaction  on  a  plate  girder  is  300,000  pounds.  If  a 
compressive  fiber  stress  of  13,000  pounds  per  square  inch  is  allowed  on  the 
stiffeners,  and  4  stiffeners  are  used,  as  shown  in  Fig.  48,  what  should 
be  the  size  of  the  angles  ? 

SOLUTION. — The  reaction  or  greatest  shear  being  300,000  pounds, 
and  the  allowable  stress  13,000  pounds  per  square  inch,  the  area  of 
stiffeners  required  must  be  300,000-=- 13,000  =  23  square  inches;  this 
sectional  area  divided  among  4  angles,  gives  23  -=-  4  =  5.75  square  inches 
as  the  area  required  for  each  angle. 

By  referring  to  the  list  of  angles  with  even  legs,  in  the  table  "Areas 
of  Angles,"  it  is  seen  that  a  5"  X  5"  X  |"  angle  has  a  sectional  area  of 
5.86,  while,  in  the  list  of  areas  of  the  uneven  angles,  a  5"  X4"  X  {j"  is 
shown  to  have  an  area  of  5.72  square  inches;  therefore,  either  of  these 
angles  may  be  used.  Ans. 


FLANGES. 

41.  The  flanges  of  a  riveted  girder  include  all  the 
metal  at  the  top  and  bottom  of  the  girder,  and  are  some- 
times called  the  top  and  bottom  chords,  though  this  term  is 
more  frequently  applied  to  lattice  or  open  girders,  such  as 
are  more  often  used  for  railroad  and  highway  bridges. 

In  building  construction,  it  is  customary  to  include  in  the 
flange  the  two  flange  angles,  the  flange  plates,  and  ^  of  the 
web-plate  included  between  the  flange  angles.  The  build- 
ing ordinances  of  some  of  the  large  cities  in  the  United 
States,  however,  disregard  this  last  item,  and  will  not  allow 
any  portion  of  the  web-plate  to  be  included  as  part  of  the 
flange.  In  this  section  the  web-plate  will  not  be  considered 
in  calculating  the  flange  area. 

If,  because  of  economic  considerations,  |  of  the  web-plate 
must  be  included  as  part  of  the  flange,  it  must  be  remem- 
bered that  the  plate  should  never  be  spliced  near  the  center, 
when  the  girder  is  uniformly  or  symmetrically  loaded,  or 
directly  under  the  point  of  greatest  bending  moment,  when 
the  load  upon  the  girder  is  unsymmetrically  placed.  Especial 
care  must  also  be  taken  to  insure  that  any  splice  made  upon 
the  length  of  such  a  web-plate  is  so  designed  as  to  furnish 


68 


ARCHITECTURAL  ENGINEERING. 


the  greatest  possible   percentage   of  strength   of  the  solid 
plate  included  within  i  of  the  depth  of  the  web. 

The  best  practice  dictates  that  where  flange  plates  are 
used,  the  sectional  area  of  the  flange  angles  should  equal  the 
sectional  area  of  the  flange  plates.  This,  however,  is  not 
possible  in  heavy  work,  where  the  best  that  can  be  done  is 
to  use  the  heaviest  sections  obtainable  for  the  flange  angles. 

4:2.  Flange  Stresses. — In  a  simple  girder  the  top  flange 
is  subjected  to  compression,  and  the  bottom  flange  to  ten- 
sion. Nevertheless,  it  is  customary  in  practice  to  make  the 
two  flanges  equal  and  composed  of  the  same  size  of  rolled 
plates  and  angles. 

In  proportioning  the  flanges  of  a  plate  girder,  the  lower 
flange  is  calculated  for  tension;  the  areas  of  the  rivet 
holes  cut  out  of  the  flanges  are  deducted  from  the  total  area, 
so  as  to  give  the  net  or  actual  area  of  the  flange  at  the  point 
of  least  strength. 

The  stresses  in  the  flanges  are  assumed  to  be  produced 
wholly  by  the  bending  moment  on  the  girder,  and  the 
moments  of  these  stresses  are  assumed  to  be  equal  to  the 
moments  of  the  external  forces. 


The  principles  on  which  the  flange  stresses  of  a  plate 
girder  are  calculated  will  be  made  clear  by  reference  to 


§  6  ARCHITECTURAL  ENGINEERING.  09 

Fig.  41>,  which  shows  a  girder  in  two  sections  joined  by  a' 
hinge  pin  i~  at  the  upper  flange,  and  a  chain  at  the  lower 
flange.  The  resultant  moment  of  the  loads  and  reactions 
tends  to  produce  rotation  about  the  center  c,  which,  how- 
ever, is  taken  at  a  point  in  the  upper  flange  instead  of  on 
the  neutral  axis,  as  was  done  in  the  case  of  the  beam  com- 
posed of  a  single  section ;  in  reality,  owing  to  the  fact  that 
the  web  is  entirely  neglected  in  calculating  the  resistance  of 
the  girder  to  the  bending  stresses,  there  is  no  neutral  axis, 
in  the  sense  in  which  that  term  was  used  in  connection  with 
ordinary  beams. 

The  stress  in  the  chain,  which  represents  the  lower  chord 
or  flange  of  the  girder,  resists  the  tendency  to  rotation  about 
the  center  of  moments  i\  a  lever  arm  /^  which  is  the  per- 
pendicular distance  from  the  chain  to  the  point  c.  It  is 
evident,  then,  that  the  strength  of  the  girder  depends  upon 
two  factors,  the  tensile  strength  of  the  lower  chord,  and 
its  distance  from  the  center  of  the  hinge  c,  the  latter  of 
which  represents  the  depth  of  the  girder. 

If  now  the  center  of  moments  is  taken  on  the  center  line 
of  the  chain,  directly  under  the  point  t\  it  is  evident  that  the 
resultant  moment  of  the  external  forces,  with  respect  to  this 
center,  is  the  same  as  when  the  center  was  taken  at  c;  it  is 
also  evident  that  the  force  in  the  beam  whose  moment,  with 
respect  to  this  center,  balances  the  resultant  moment  of  the 
external  forces,  is  the  compression  on  the  pin  c.  Since  the 
moment  and  the  lever  arm  of  the  compressive  stress  on  the 
pin  are  respectively  equal  to  the  moment  and  the  lever  arm 
of  the  tensile  stress  in  the  chain,  it  follows  that  these  two 
stresses  are  equal;  in  other  words,  the  compressive  stress  in 
the  top  flange  of  the  girder  is  equal  to  the  tensile  stress  in 
the  bottom  flange. 

43.  Proportioning?  the  Flanges. — Having  determined 
the  principles  upon  which  the  bending  strength  of  a  plate 
girder  is  calculated,  it  remains  to  show  a  method  for  propor- 
tioning the  metal  in  the  flanges.  The  usual  process  is  as 
follows : 


70  ARCHITECTURAL  ENGINEERING.  §  6 

•  First  calculate  the  maximum  bending  moment  upon  the 
girder;  this  may  be  done  by  the  principles  and  rules  given 
in  Architectural  Engineering,  §  5.  In  calculating  the  bend- 
ing moment  on  a  plate  girder,  however,  it  is  customary  to 
express  the  moment  in  foot-pounds,  the  depth  of  a  girder 
being  generally  given  in  feet  and  not  in  inches,  as  in  solid 
beams  of  shallow  depth.  If,  however,  the  depth  of  the 
girder  is  expressed  in  inches,  the  bending  moment  must  be 
calculated  in  inch-pounds. 

Having  found  the  maximum  bending  moment  on  the 
girder,  it  is  necessary  to  assume  an  allowable  fiber  stress  for 
the  material  of  which  the  flanges  are  composed.  The  fol- 
lowing rule  may  then  be  used  to  calculate  the  sectional  area 
of  either  flange. 

Rule.  —  Divide  the  bending  moment  on  the  girder  in  foot- 
pounds by  the  product  obtained  by  multiplying  the  depth  of 
the  girder  in  feet  by  the  safe  fiber  stress. 

The  safe  fiber  stress  for  a  given  case  is  obtained  by 
dividing  the  ultimate  fiber  stress  per  square  inch  of  the 
material  by  the  factor  of  safety  required  in  the  girder. 

The  rule  may  "be  expressed  by  the  formula 


in  which  A  =  net  area  of  one  flange  in  square  inches; 
D  =  depth  of  the  girder  in  feet  ; 
vS  =  safe   fiber   stress   per    square    inch    of    the 

material; 

M  =  bending   moment    on    the    girder   in    foot- 
pounds. 

EXAMPLE.—  The  depth  of  a  plate  girder  is  6  feet,  the  span  is  80  feet, 
and  the  load  upon  the  girder  is  3,000  pounds  per  lineal  foot,  (a)  What 
will  be  the  required  net  flange  area  for  structural  steel,  if  a  factor  of 
safety  of  4  is  used  ?  (b)  Of  what  size  rolled  sections  should  the  flange 
be  composed  ? 

SOLUTION.—  The  span  being  80  feet  and  the  load  3,000  pounds  per 
lineal  foot,  the  entire  load  on  the  girder  will  be  80  X  3,000  =  240,000 
pounds. 


ARCHITECTURAL  ENGINEERING. 


71 


Substituting  in  the  formula  J/  =  -^-^  for  the  bending  moment  on 

a  simple  beam  (see  Table   10,  Art.  97,  Architectural  Engineering, 
§  5),  we  have 


M  = 


240,000  X  BO 


=  2,400,000  foot-pounds. 


The  net  area  of  the  flange,  from  formula  14,  is 
2,400,000 


A  = 


6X15,000 


-3 X/4  Flange  Plates 


ngles. 


All  Rivets  gat/am. 


(b]  In  order  to  determine  the  size  of  the  flange  plates  and  angles,  it 
is  useful  to  assume  some  particular 
size  of  angle  and  plates,  and  make 
a  detail  sketch  of  the  flange,  as 
shown  in  Fig.  50,  marking  on  it  the 
size  of  the  respective  plates  and 
angles  that  have  been  assumed. 
The  rivets  should  also  be  shown, 
so  that  the  metal  cut  out  of  the 
rivet  holes  may  be  deducted  from 
the  sectional  area  of  the  flange  in 
order  to  determine  that  area. 

It  is  assumed  in  the  section 
under  consideration  that  there  are 
two  rows  of  rivets  through  the  ver- 
tical legs  of  the  angles,  each  pair  of  these  rivets  being  placed  in  the 
same  vertical  plane  in  consequence  of  which  the  amount  to  be  deducted 
from  the  net  section  is  double  the  area  cut  out  for  one  rivet.  The  rivets 
in  the  two  rows  through  the  horizontal  legs  of  each  angle  are  staggered, 
and  consequently  only  one  rivet  hole  in  each  horizontal  leg  affects  the 
area  of  the  flange  section. 

According  to  the  table  "Areas  of  Angles,"  the  area  of  a  6"  X  6"  X  f" 
angle  is  7.11  square  inches,  therefore,  the  total  area  of  the  metal  in  the 
flange  is 

2-6"  X  6"  X  f"  angles,  7.11  X  2  =  1  4.2  2  sq.  in. 

4-14"  X  |"  plates,      4  X  14  X  f  =  2  1.0  0  sq.  in. 

Total, 3  5.2  2  sq.  in. 


FIG.  50. 


From  the  total  area  of  the  flange  it  is  necessary  to  deduct  the  metal 
cut  out  for  the  rivet  holes.  As  |-inch  rivets  are  used,  the  rivet  holes 
are  considered  to  be  \  of  an  inch  larger,  or  1  inch  in  diameter. 

There  are  four  1-inch  holes  through  |  inch  of  metal  to  be  deducted 
from  the  vertical  legs  of  the  angles,  and  in  the  plates  and  the  horizontal 


ARCHITECTURAL  ENGINEERING. 


G 


legs  of  the  angles  there  are  two  1-inch  holes  through  2£  inches  of  metal. 
The  areas  to  be  deducted  for  the  rivet  holes  are,  therefore, 

4-1-inch  holes  through  f  in.  of  metal     =  2J  sq.  in. 
3-1-inch  holes  through  2£  in.  of  metal  =  4£  sq.  in. 

Total, 6f  sq.  in. 

and  the  net  area  of  the  flange  is  35.22-6.75  =  28.47  square  inches. 
Since  the  calculations  showed  that  the  net  area  required  in  this  flange 
is  26.6  square  inches,  it  is  evident  that  the  assumed  flange  is  amply 
strong.  Ans. 

44.  Lengths  of  Flange  Plates. — Since  the  bending 
moment  in  a  simple  beam  varies  along  the  entire  length 
of  the  beam,  the  location  of  the  maximum  bending  moment 
depending  on  the  distribution  of  the  load,  it  would  seem 
that,  in  order  to  design  an  economical  girder,  the  area  of  the 
flange  should  vary  with  the  bending  moment.  Where  flange 
plates  are  used,  this  condition  may  be  partially  fulfilled  by 
the  use  of  plates  of  different  lengths,  each  extending  only  as 
far  as  may  be  required  in  order  to  provide  the  flange  section 
demanded  by  the  bending  moment. 

Reference  to   Fig.  51  will  make   this  construction   more 

Top  Plate 


FIG.  51. 

clear.  In  this  figure,  it  is  seen  that  the  topmost  plate  of  the 
top  flange  is  the  shortest,  and  extends  over  a  small  portion 
of  the  girder  only,  each  successive  plate  under  this  one 
being  longer  than  the  one  above  it.  The  third  plate  from 
the  top  is  the  longest  and  extends  nearly  the  full  length  of 
the  girder,  while  the  angles  extend  from  end  to  end. 

Where  the  beam  is  uniformly  loaded,  the  following  method 
may  be  used  to  obtain  the  theoretical  length  of  each  of  the 
flange  plates: 

Commencing  with  the  outside  plate  of  the  flange,  find  the 


6 


ARCHITECTURAL  ENGINEERING. 


sum  of  all  the  net  areas  in  square  inches  of  the  plates  to  and 
including  the  plate  in  question.  Thus,  in  Fig.  52,  if  it  be  re- 
quired to  obtain  the  length 
of  the  third  plate  from  the 
outside,  find  the  sum  of  the 
areas  of  the  first,  second,  and 
third  plates.  If  the  length 
of  the  second  plate  is  re- 
quired, then  the  sum  of  the 
areas  of  the  first  and  second 
plates  is  to  be  taken.  Divide 
the  area  so  obtained  by 
the  net  area  of  the  whole 
flange  in  square  inches, 
and  multiply  the  square 
root  of  this  quotient  by  the  length  of  the  girder  in  feet; 
the  product  will  be  the  theoretical  length  of  the  plate  in 
feet. 

Having  obtained  the  theoretical  length  of  the  plate,  it  is 
•necessary  to  add  from  12  to  10  inches  to  each  end,  in  order 
that  the  plate  in  question  may  be  carried  sufficiently  past 
the  point  of  bending  moment  which  governs  the  area  of  the 
flange  at  its  ends  to  be  securely  riveted  to  the  plates  and 
angles  making  up  the  flange  from  there  on  to  the 
abutment. 

The  method  for  determining  the  length  of  flange  plates 
where  the  beam  is  uniformly  loaded  may  be  expressed  by 
the  formula 


FIG.  52. 


/=  L 


(15.) 


in  which  /  =  theoretical    length    in    feet    of    the    plate    in 

question ; 

L  =  the  length  of  the  girder  in  feet; 
a  —  net  area  of  all  the  plates  to  and  including  the 

plate  in  question,  beginning  with  the  outside 

plate ; 
A  =  total  net  area  of  the  entire  flange. 


ARCHITECTURAL  ENGINEERING. 


6 


Angles 


Alt  Rivets  3  'at/am. 


EXAMPLE. — In  Fig.  53  is  shown  a  section  through  the  flange  of  a 

plate  girder  the  span  of  which  is 
60  feet.  What  is  the  theoretical 
length  of  each  of  the  three  flange 
plates  ? 

SOLUTION. — Area  of  a  4"  X  4"  X  \" 
angle  according  to  the  table  "Areas 
of  Angles,"  is  3.75  square  inches. 
The  area  of  each  plate  is  f  X 12 
=  4.5  square  inches.  The  diameter 
to  be  deducted  for  the  rivet  Jioles  is 
|  +  ^  =  |  inch. 

The  area  cut  out  by  a  f-inch  hole 
through  a  f-inch  plate  is  .  875  X- 375 
=  .328  square  inch.  Then,  as  there 
are  two  rivet  holes  in  each  plate,  its  net  area  is  4.5  sq.  in.  —  (.328  sq.  in. 
X  2)  =  3.844  square  inches. 

The  net  area  of  the  angles  is  (3.75  sq.  in.  X  2)  —  (.4375  sq.  in.  X4) 
=  5.75  square  inches. 
The  net  area  of  the  flange  section  is,  therefore, 

3  plates  3.844  X  3  =  1  1.5  3  2  sq.  in. 

2  angles  =      5.7  5      sq.  in. 

Total,     .     .        1  7.2  8  2  sq.  in. 


Fio.  53. 


Now  calculate  the  length  of  the  outside  plate, 
formula,  we  have 


Substituting  in  the 


/  =  GO* 


=  28.29  feet. 


By  substituting  the  proper  values,  the  theoretical  length  of  the  sec- 
ond or  middle  plate  is 


/= 


=  40.0  feet. 


The  length  of  the  third  or  last  plate  in  the  flange,  that  is,  the  one 
next  to  the  flange  angles,  is  next  to  be  calculated,  though  some  engi- 
neers prefer  to  run  this  the  entire  length  of  the  girder,  as  it  stiffens 
the  girder  laterally  and  assists  in  preventing  any  tendency  towards 
side  deflection.  The  theoretical  length  of  this  plate  is 


=   60|/: 


11.532 
17T282 


=  49.12  feet. 


45.  Graphical  Method  of  Determining  Length  of 
Flange  Plates.— The  graphical  method  for  determining 
the  theoretical  length  of  flange  plates  in  built-up  girders  is 


6 


ARCHITECTURAL  ENGINEERING. 


F/angre  Plates. 


&-6JT6  "xjfr  Angles. 


All  Rivets j  a/iam. 


more  convenient  than  the  analytical  method  previously 
given.  In  order  to  explain  this  method,  we  will  assume  a 
section  through  the  flange  of  a  plate  girder,  and  find  the 
lengths  of  the  several  flange  plates. 

Fig.  54  shows  a  section  through  the  flange  of  a  girder, 
built  up  of  four  \"  X 14" 
flange  plates,  the  span  of 
the  girder  being  90  feet. 
It  will  be  noticed  that 
there  are  two  rows  of 
rivets  in  the  flange,  and 
twro  rows  in  the  vertical  leg 
of  the  angles,  but  as  the 
latter  are  staggered,  there 
will  be  but  one  rivet  hole 
to  be  deducted  from  the 
vertical  leg  of  each  angle. 

The  sectional  area  of  a  6"x6"xf"  angle  is  found  by  the 
table  "Areas  of  Angles"  to  be  8.44  square  inches;  from 
this  deduct  1|-  square  inches,  the  area  cut  out  by  the  two 
rivet  holes,  making  the  net  area  of  each  flange  angle  G.94 
square  inches. 

The  sectional  area  of  a4-"x!4"  flange  plate  is  7  square 
inches,  from  which  there  is  to  be  deducted  1  square  inch  for 
the  sectional  area  cut  out  by  the  two  rivet  holes.  Hence, 
the  net  area  of  one  flange  plate  is  7  sq.  in.— 1  sq.  in.  =  (J 
square  inches. 

The  net  area  of  the  entire  flange  will,  therefore,  be 
2-6"x6"xf"  angles        =  1  3.8  8  sq.  in. 
4-|"xl4"  flange  plates  =  2400  sq.  in. 
Total,     .      .      .        3  7.8  8  sq.  in. 

We  will  now  proceed  with  the  diagram,  see  Fig.  55. 
Since  the  load  is  uniformly  distributed,  the  flange  plates 
extend  equally  on  each  side  of  the  center;  consequently, 
the  diagram  for  only  one-half  of  the  girder  will  be  drawn. 
Draw,  to  any  scale,  a  horizontal  line  a  b,  Fig.  55,  equal  to 
one-half  of  the  span;  divide  this  line  into  any  number  of 


7(J 


ARCHITECTURAL  ENGINEERING. 


equal  parts   (in  the  figure,  twelve  parts  have    been  used). 
Upwards    from   the    points   of    division,    draw    indefinite 


§  G  ARCHITECTURAL  ENGINEERING.  77 

perpendicular  lines.  On  the  perpendicular  from  b,  lay  off 
to  some  scale  a  distance  which  represents  the  entire  net 
section  of  the  flange,  thus  locating  the  point  r.  For 
example,  the  net  area  of  the  flange  in  this  case  is  37.88 
square  inches;  letting  TL  inch  represent  1  square  inch,  the 
distance  br  must  be  37.88X7^  =  2.37  inches,  nearly. 

Lay  off  to  the  same  scale  on  the  line  /;  r  a  distance  b  n, 
which  represents  13.88  square  inches,  the  net  area  of  the 
two  flange  angles;  also  the  distances  no,  op,  pq,  and  qr, 
each  representing  G  square  inches,  the  net  area  of  each  of 
the  flange  plates.  From  the  point  r,  draw  a  horizontal  line 
cutting  the  vertical  line  erected  at  a,  thus  locating  the 
point  b'.  Divide  the  vertical  line  a  b'  into  the  same  number 
of  equal  parts  as  the  line  a  b,  thus  locating  the  points  c',  </', 
c',f,  etc.;  and  from  these  points,  draw  the  lines  c' r,  d'  r, 
etc.  Draw  the  curve  a  si'  r  through  the  points  where  the  ver- 
tical lines  from  c,  tf,  c,  etc.  intersect  the  corresponding  lines 
c' r,  d'  r,  c'  r,  etc.  Now  from  the  points  q,  /,  o,  and  //,  draw 
horizontal  lines  as  shown,  cutting  the  curve  in  the  points 
s,  t,  K,  and  i',  and  from  each  of  these  points  of  intersection, 
draw  a  perpendicular,  extending  it  imtil  it  intersects  the 
horizontal  line  next  above.  The  rectangles  i'' r  q  \\  u'qpu, 
t'pot,  etc.,  thus  formed,  represent  the  flange  plates 
and  angles. 

Now  in  order  to  obtain  the  theoretical  length  of  any 
flange  plate,  measure  the  length  of  the  corresponding 
rectangle  by  the  scale  to  which  the  half  span  was  laid  out 
on  the  line  ab;  this  length  multiplied  by  2  gives  the  length 
of  the  plate  in  question.  For  example,  if  it  is  desired  to 
obtain  the  length  of  the  first,  or  top,  plate,  with  the  scale  to 
which  the  half  span  was  laid  out,  measure  the  length  of  the 
line  v1 >;  as  only  one-half  of  the  diagram  is  drawn,  this 
gives  one-half  of  the  length  of  the  top,  or  first,  plate,  and  by 
doubling  this  the  entire  theoretical  length  of  the  plate  in 
question  is  obtained.  The  length  of  the  other  plates  may 
be  determined  in  like  manner.  It  is  to  be  borne  in  mind 
that  to  the  theoretical  length  given  by  the  diagram  it  is 
necessary  to  add  a  length  of  aboiit  1  foot  at  each  end  of  the 


78 


ARCHITECTURAL  ENGINEERING. 


6 


plate.  From  the  diagram,  the  theoretical  lengths  of  the 
flange  plates  of  the  girder  shown  in  Fig.  54  are  found  to  be 
35  ft.  2  in.,  50  ft.  5  in.,  62  ft.  2  in.,  and  71  ft.  11  in., 
respectively. 

The  student  will    find  upon  checking  these   lengths  by 
formula  15  that  they  are  approximately  correct.     Fig.  56, 

Upon  Sca/ina  fn&se  Distances  with  the  same  •Scats  if  which  the  I 

I  Length  o/fhe  Linedb  was  measured,  the  theoref/ca/ Length   \ 

f*    of  tf/af'/anoe  P/ates  may  be  oofaitnd. 


Divide  this  Distance  into  any  number  of  equal  Paris  — 


i.  ay  qff  this  Distance  with  any  con  venientSca/e,  - 

eaua/  to  One  Ha/ f  the  Span  of  the  Sirdet: 


3 


FIG.  56. 


in  which  all  the  different  steps  are  indicated,  is  presented  in 
order  that  the  student  may  always  have  a  guide  for  laying 
out  this  diagram. 

46.  Application  of  the  Graphical  Method  to  Girders 
With  Concentrated  Loads. — The  graphical  method  for 
determining  the  theoretical  length  of  flange  plates  when  the 
girder  is  loaded  with  concentrated  loads,  which  is  similar  to 
that  given  for  a  uniformly  loaded  girder,  will  be  illustrated 
by  constructing  a  diagram  for  the  lengths  of  the  four  flange 
plates  required  for  a  girder  with  a  span  of  80  feet  and  a 
depth  of  6  feet,  carrying  a  concentrated  load  of  185,000 
pounds  at  30  feet  from  one  end. 

The  bending  moment  on  the  girder  may  be  calculated  by 


§  G  ARCHITECTURAL  ENGINEERING.  70 

the  method  given  in  Architectural  Engineering,  §5,  or  by 
the  formula  / 


in  which  J/  =  bending  moment  ; 
W  =  load  on  the  girder; 
L   =  span; 
a    —  distance  that   the  load  is  located   from   one 

abutment; 
b    =  distance  it  is  located  from  the  other. 

In  Fig.  57  it  is  seen  that  the  load  is  located  at  the  point  f 
30  feet  from  R^  and  50  feet  from  R^  Substituting  these 
values  in  formula  16,  the  bending  moment  is  found  to  be 


M  —  185,OOOX—  =  3,  468,750  foot-pounds. 

80 

From  this  bending  moment  the  required  net  flange  area, 
assuming  a  safe  unit  stress  of  15,000  pounds  per  square 
inch,  is  found  to  be,  approximately,  38  square  inches,  which 
is  provided  by  the  use  of  four  £"xl4"  plates  and  two 
6"X6"X%"  angles.  The  flange  therefore  has  the  section 
shown  in  Fig.  54. 

To  construct  the  diagram,  which  is  shown  in  Fig.  57, 
draw  the  base  line  c  d  to  any  scale  equal  to  the  span  of  the 
girder  in  feet.  (In  this  case,  owing  to  the  fact  that  the  load 
is  not  symmetrically  placed,  the  center  line  of  the  girder 
will  not  divide  the  lengths  of  the  plates  in  halves,  and  it 
will  be  necessary  to  draw  the  entire  diagram.)  Locate 
the  point  of  application  of  the  concentrated  load  at  30  feet 
from  R^  and  draw  the  perpendicular  line  cf.  On  this  line 
lay  off  to  any  scale  a  distance  which  represents  the  bending 
moment.  For  example,  in  this  case,  a  scale  has  been  used 
on  which  50,000  foot-pounds  is  represented  by  -^  of  an  inch; 
the  distance  to  be  laid  off  is,  therefore,  *$«*$*a.xjs  =  2.108 
inches.  This  locates  the  point  g,  wrhich  is  then  connected 
by  straight  lines  to  the  points  c  and  d.  Draw  vertical  lines 
from  the  points  c  and  d,  until  they  meet  a  horizontal  line 
through  the  point  g,  at  q  and  r. 


80 


ARCHITECTURAL  ENGINEERING. 


In  the  flange  section,  the  angles  have  a  combined  net  area 
of  13.88  square  inches,  and  each  flange  plate  a  net  area  of 


6  square  inches.     The  combined  net  sectional  area  of  the 
flange  is  37.88  square  inches. 

Place  the  zero  mark  of  any  convenient  scale  at  the  point 


§  0  ARCHITECTURAL  ENGINEERING.  81 

/,  and  slant  the  scale  until  the  marking  on  the  scale  which 
represents  37.88  square  inches,  the  net  area  of  the  flange, 
falls  upon  the  horizontal  line  q  r.  Thus,  in  this  case,  a 
TL-inch  scale  has  been  used,  and  the  zero  mark  is  placed  at 
f,  while  the  mark  on  the  scale  representing  the  division 
37.88  is  placed  on  the  horizontal  line  q  r.  With  the  scale 
still  in  this  position,  begin  at  the  zero  mark  and  lay  oft"  a 
distance  13.88  to  represent  the  net  sectional  area  of  the  two 
angles;  then  lay  off  four  distances,  each  equal  to  ti,  to  repre- 
sent the  net  sectional  area  of  each  of  the  flange  plates. 

Through  the  points  thus  found,  draw  the  horizontal 
lines  // /,  /'/,  /////,  and  op.  At  the  points  suicy,  etc., 
where  these  horizontal  lines  cut  the  oblique  lines  gc  and 
gd,  draw  the  perpendicular  lines  st,  u  :•,  ic x,  r  z,  a'  b',  etc.. 
extending  each  until  it  meets  the  next  horizontal  line  above. 
Then  the  rectangles  enclosed  by  the  horizontal  and  vertical 
lines,  shown  heavy  in  the  diagram,  represent  the  cover- 
plates  and  flange  angles.  By  measuring  the  length  of  these 
rectangles  with  the  scale  to  which  the  span  was  laid  off  on 
the  line  f  <•/,  the  theoretical  length  of  the  plates  may  be 
determined. 

In  this  case  the  length  of  the  angles  is  equal  to  the  span, 
80  feet,  as  marked  on  the  diagram.  The  length  of  the  first, 
or  top,  plate  measures  12  ft.  0  in.,  of  the  second  plate  20  ft. 
0  in.,  of  the  third  39  ft.  3  in.,  and  of  the  fourth,  or  last,  52  ft. 
0  in.  It  may  be  that  when  the  student  lays  out  the  diagram 
for  himself,  he  will  obtain  results  which  will  vary  slightly 
from  those  given.  However,  a  variation  of  a  fractional  part 
of  an  inch  in  the  length  of  a  flange  plate  on  a  girder  need 
not  be  considered. 

47.  Graphical  Diagram  for  Several  Concentrated 
Tjoads.— In  order  to  illustrate  this  method  still  further,  a 
more  complicated  problem,  in  which  there  are  three  con- 
centrated loads,  will  now  be  presented. 

Assume  that  a  girder  having  a  span  of  80  feet  with  a 
depth  of  6  feet  is  loaded  as  shown  in  Fig.  58.  What  flange 
area  is  required,  provided  a  safe  unit  fiber  stress  of  1G,0<>0 

1-27 


82  ARCHITECTURAL  ENGINEERING.  §  6 

pounds  per  square  inch  is  used,  and  of  what  should  the  flange 
be  constructed,  also  what  will  be  the  length  of  the  several 
flange  plates  ? 

The  reactions  at  Rl  and  Ra  are  found  to  be  142,500  and 
117,500  pounds,  respectively.     It  will  first  be  necessary  to 


•^-Loadb-eOOOV/b. 

-  —  Loac/c-l£0000lb. 

•^-—so'-o"-  » 

Loada-8OOOO/t>.  -~ 

§ 

^_ 

^ 

FIG.  58. 

calciilate  the  bending  moment  in  foot-pounds  at  the  points 
a,  b,  and  c,  where  the  loads  are  concentrated. 

The  bending  moment  at  a  is  142,500x20  =  2,850,000 
foot-pounds;  at  b  the  bending  moment  is  (142,500x30) 
-(80,000X10)  =  3,475,000  foot-pounds;  and  the  bending 
moment  at  c  is  (142,500x50)  -[(80,000x30)  +  (60,000 
X20)]=  3,525,000  foot-pounds. 

From  the  above  it  is  seen  that  the  greatest  bending 
moment  is  under  the  load  located  at  c,  and  its  magnitude 
is  3,525,000  foot-pounds. 

From  this  the  flange  area  required  may  be  calculated  by 
applying  formula  14,  as  follows: 
3,525,000 


A  = 


=  36.71  square  inches. 


6X16,000 

We  will  now  select  a  flange  section  which  will  have  the 
required  net  sectional  area ;  referring  to  the  section  shown 
in  Fig.  54,  the  area  is  found  to  be  37.88  square  inches;  this 
flange  will  therefore  satisfy  the  requirements. 

Having  determined  the  bending  moments  and  the  great- 
est net  flange  area,  begin  the  diagram  shown  in  Fig.  59  by 
drawing  to  any  scale  the  horizontal  line  de  equal  in  length 
to  the  span  of  the  girder ;  with  the  same  scale  locate  the 


§  0  ARCHITECTURAL  ENGINEERING.  S3 

points  of  application  a,  b,  and  c  of  the  concentrated  loads. 
Upwards  from    the    points  d,  a,  b,  c~,  and  <•,   draw  indefinite 


/ 

[ 

/ 

"g 

. 

; 

/ 

x7 

r- 

--.  \ 
<-;  ^ 

s 

i  ^ 

s  ; 

, 
•:  r 
>  « 

< 

i) 
) 

55 

s 

s 

1 

, 







\ 

L 



-- 

s, 

\ 

perpendicular  lines,  and  on  the  perpendiculars  from  a,  b, 
and  c,  lay  off  to  some  convenient  scale  distances  a/,  bg,  and 


ARCHITECTURAL  ENGINEERING. 


§G 


c  h,  which  represent  the  respective  bending  moments  at  these 
points. 

For  example,  the  bending  moment  at  a  is  2,850,000  foot- 
pounds; at  b  it  is  3,475,000  foot-pounds;  and  at  c  it  is 
3,525,000  foot-pounds;  therefore,  assuming  a  scale  on  which 
each  -jV  of  an  inch  represents  50, 000  foot-pounds  of  bending 
moment,  the  respective  bending  moments  at  the  points  a,  6, 
and  c  are  represented  by  lengths  af  of.  57  thirty-seconds,  bg 
of  69|  thirty-seconds,  and  c  Ii  of  70^  thirty-seconds. 

Draw  straight  lines  connecting  the  points  d,/,g,  //,  and  e. 

L BySca/inatheseDistances,withthe j 

same  Sca/e  to  which  the  L  er?afh  of  the 


Line  db  was  laid  off,  the  theoretical  • 

~~   Length  of  the  F/ange  P/ates 


L  ciy  off  this  Distance  with  any  convenient  Scale 
to  e<pua//  the  Sf>an  of  the  Gt  refer. 

t« —  With  the  same  Scale,  locate  thepo/nts  of  ajbf>/icat/on 

ofth&  concentrated  Loads,as  at  c,d,e. 

FIG.  60. 

Through  the  highest  point  //,  representing  the  greatest 
bending  moment,  draw  the  horizontal  line  j  k. 

The  net  area  of  the  flange  being  37.88  square  inches, 
place  the  zero  mark  of  any  convenient  scale  on  the  line  de, 
and  slant  the  scale  until  the  mark  which  represents  37.88 
falls  on  the  line  j  k. 

Starting  from  the  zero  mark  on  the  scale,  lay  off  a  dis- 
tance which  represents  the  net  area  of  the  two  flange  angles, 
in  this  case  13.88  square  inches,  then  divide  the  remaining 
distance  into  equal  parts,  each  of  which  represents  6  square 
inches,  the  net  sectional  area  of  the  several  flange  plates. 


ARCHITECTURAL  ENGINEERING. 


85 


Through  the  points  just  found,  draw  the  horizontal  lines 
/;;/,  no,  pq,  and  r s.  Where  these  horizontal  lines  intersect 
the  oblique  lines  at  the  points  /,  //,  i\  ic,  .r,  etc.,  draw  ver- 
tical lines  until  they  intersect  the  next  horizontal  line  above. 
Then  draw  in  with  heavy  lines  the  rectangles  representing 
the  flange  plates  and  flange  angles. 

The  theoretical  length  of  the  flange  plates  may  now  be 
determined  by  measuring  with  the  scale  to  which  the  span 
was  laid  off  on  the  line  d  c. 

The  length  of  the  top,  or  first,  plate  in  this  case  is  found 
to  be  32  ft.  11  in. ;  of  the  second  plate,  42  ft.  0  in.  ;  of  the 
third  plate,  51  ft.  3  in. ;  and  of  the  fourth,  or  last,  plate, 
59  ft.  9  in. 

In  Fig.  60  is  shown  a  diagram  which  will  serve  as  a  gen- 
eral rule  for  determining  the  length  of  the  several  flange 
plates  of  a  girder  loaded  with  several  concentrated  loads. 

48.  Diagram  for  a  Combination  of  Concentrated 
Loads  With  a  Uniformly  Distributed  Load. — There  is 
still  another  condition  of  girder  loading  which  is  frequently 
encountered  in  practical  work,  and  in  which  it  is  necessary 
to  determine  the  length  of  the  several  flange  plates  by  the 


ao'-o" 


L  oadb~45OOOO  /t>. 


.  f 


Spar?  6O  Feet    - 

FIG.  61. 

graphical  method ;  this  condition  is  produced  by  a  combi- 
nation of  a  uniformly  distributed  load,  with  several  concen- 
trated loads  located  at  different  points  along  the  girder. 

In  order  to  explain  the  method  for  obtaining  the  length  of 
the  flange  plates  in  a  girder  loaded  in  this  manner,  the 
following  problem  will  be  assumed  and  the  diagram  will  be 
constructed  as  was  done  in  previous  cases. 


86 


ARCHITECTURAL  ENGINEERING. 


4-ixi3  F/artae  Plates 


Assume  the  girder  to  be  loaded,  as  shown  in  Fig.  61,  with 
a  uniformly  distributed  load,  and  the  two  concentrated  loads. 
The  flange  section  shown  in  Fig.  62  is  sufficient  to  resist  the 

bending  moments  due  to  these 
loads.  It  is  required  to  deter- 
mine by  the  graphical  method 
the  theoretical  lengths  of  the 
a-6x6'xj"Any/es  several  cover-plates  making  up 
the  flange  section. 

Before  starting  to  draw  the 
diagram  shown  in  Fig.  63,  it  is 
necessary  to  make  the  calcula- 
tions for  the  following:  The 
greatest  bending  moment;  the 
maximum  bending  moment  due 
to  the  uniformly  distributed 
load  ;  and  the  bending  moment  under  each  of  the  concen- 
trated loads,  neglecting  the  uniformly  distributed  load. 
These  bending  moments  should  be  expressed  in  foot-pounds. 


m.  Rivets, 


FIG.  62. 


FIG.  63. 


The  flange  area  required  to  successfully  resist  each  of  these 
bending  moments  should  also  be  calculated. 


§  G  ARCHITECTURAL  ENGINEERING.  87 

The  calculations  in  this  case  have  been  made  in  the  usual 
manner,  and  the  results  are  as  follows: 

Greatest  bending  moment =  2,070,000  ft.-lb. 

Bending  moment  due  to  a  uniform  load    =      1)00,000  ft.-lb. 
Bending    moment   under   concentrated 

loadrt =  1,170,000  ft.-lb. 

(Considering  the  concentrated  loads  only.) 
Bending    moment   under   concentrated 

load  b =      792,000  ft.-lb. 

(Considering  the  concentrated  loads  only.) 

Since  the  depth  of  the  girder  is  4  feet,  if  a  unit  fiber  stress 
of  15,000  pounds  is  used;  the  flange  area  required  to  resist 
the  greatest  bending  moment  is 

2,070,000 

A  =   ,     ,  ^   „  „  =  34.o  square  inches. 
4xlo,000 

The  flange  area  required  to  resist  the  bending  moment 
due  to  the  uniform  load  is 

900,000 

A  =  -  —  =  lo  square  inches. 

4x15,000 

The  flange  area  required  to  resist  the  bending  moment 
at  the  point  on  the  girder  where  the  concentrated  load 
a  is  situated,  considering  the  concentrated  loads  only,  is 

1,170,000 
A'-~-  4^I5TOOO  ==!»*  square  mches. 

The  flange  area  required  to  resist  the  bending  moment  at 
the  point  on  the  girder  under  the  concentrated  load  b  is 

792,000 
:  4X15>Q()0  -  13.2  square  inches, 

considering  as  before  only  the  concentrated  loads, 

As  the  loads  on  the  girder  are  not  symmetrically  placed 
with  regard  to  the  center,  it  will  be  necessary  to  draw  the 
complete  diagram.  Begin  the  diagram  by  drawing  to  any 
convenient  scale  the  horizontal  line  c  d,  Fig.  63,  equal  in 
length  to  the  span  of  the  girder,  and  locate  the  points  of 
application  of  the  concentrated  loads  at  a  and  b\  upwards 
from  the  points  c,  a,  b,  and  d,  draw  indefinite  vertical  lines. 


88  ARCHITECTURAL  ENGINEERING.  §  6 

Now  in  accordance  with  the  method  explained  in  Art. 
45,  make  the  construction  to  determine  the  curved  line 
representing  the  bending  moment  due  to  the  uniform  load, 
as  follows : 

At  the  center  of  the  girder  draw  a  vertical  line  (in  this 
case  the  center  of  the  girder  is  found  to  be  at  the  point  where 
the  load  a  is  concentrated) ;  divide  half  the  span  into  any 
number  of  equal  parts,  as  at  e,  f,  g,  //,  z,  etc. ,  and  from  the 
points  so  obtained  draw  perpendiculars.  Lay  off  on  the 
vertical  line  passing  through  the  center  a  distance  a  /,  which 
may  represent  either  the  greatest  bending  moment  at  this 
point  due  to  the  uniform  load,  or  the  flange  area  required  to 
resist  this  bending  moment,  as  they  are  proportional.  In 
this  case  the  net  area  required  in  the  flange  will  be  used ; 
hence,  as  the  area  of  the  flange  required  for  the  uniform  load 
is  15  square  inches,  if  -^  of  an  inch  is  assumed  to  represent 
1  square  inch  of  flange  area,  the  distance  a  I  will  be  |£  inch. 

Through  the  point  /,  draw  the  horizontal  line  m  n.  Divide 
the  distance  me  into  the  number  of  equal  parts  that  the 
half  of  the  span  was  divided  into,  and  from  the  points  /*,  j, 
/,  ?/,  i',  etc.  thus  obtained,  draw  converging  lines  to  the 
point  /;  where  these  oblique  lines  intersect  the  vertical 
lines,  mark  the  points  a',  b ',  c',  d' ,  etc.,  and  through  these 
points  draw  the  curve  c  /.  Draw  the  other  half  of  the  curve 
Id  in  the  same  manner,  thus  completing  the  diagram  for 
the  uniform  load. 

Now  draw  the  diagram  for  the  concentrated  loads.  On 
the  vertical  line  a  /,  extended,  lay  off  the  distance  a  /if, 
equal  to  the  net  flange  area  required  to  support  the  concen- 
trated load  a,  and  on  the  perpendicular  line  erected  at  #,  lay 
off  the  distance  b  i'  equal  to  the  flange  area  required  at  the 
point  b  to  support  the  concentrated  loads. 

It  must  be  remembered  that  the  same  scale  is  to  be  used 
as  that  with  which  the  flange  area  required  for  the  uniform 
load  diagram  was  laid  off ;  also  that  if  the  vertical  distances 
are  laid  off  to  represent  the  bending  moment  in  the  one 
case,  the  bending  moment  should  be  used  in  the  other, 
while  if  the  flange  area  required  is  used  in  the  one  case,  it 


§  G  ARCHITECTURAL  ENGINEERING.  W) 

is  self-evident  that  it  should  also  be  used  in  the  other.  The 
student  will  understand  the  importance  of  the  above  when 
he  proceeds  further  with  the  diagram. 

Having  located  the  points  h'  and  /',  connect  with  straight 
lines  the  points  c,  h',  /',  and  d,  as  was  done  in  the  previous 
similar  diagrams  of  concentrated  loads.  This  completes  the 
diagram  for  the  concentrated  loads. 

The  next  step  in  the  process  is  to  measure  the  distance  cj" 
with  a  pair  of  dividers,  and  from  the  point  a'  on  the  curve 
representing  the  uniform  load,  lay  off  on  the  vertical  line 
the  distance  a' k'  equal  to  cj',  also  lay  off  from  the  point  // 
the  distance  b1 in'  equal  to/"/',  and  from  the  point  c'  lay  off 
on  the  vertical  line  the  distance  c'  H'  equal  to  go'  \  continue 
in  this  manner  through  the  entire  diagram.  Having  in  this 
manner  determined  the  points  /•',  in',  >i',f>',  etc.  through 
the  entire  diagram,  draw  in  the  curve  c  k'  in'  n' ,  etc.  The 
point  /'  is  the  highest  point  in  the  diagram,  and  its  distance 
from  the  horizontal  line  c  d  represents  the  entire  flange  area 
required  in  the  girder  to  resist  the  greatest  bending  due  to 
both  the  uniform  and  the  concentrated  loads. 

Through  the  point  /'  draw  the  horizontal  line  it'?1',  and 
lay  off  between  the  horizontal  lines  u'  v'  and  c  d  the  several 
distances  representing  the  net  area  of  the  flange  plates  and 
flange  angles.  Through  the  points  of  these  divisions,  draw 
the  horizontal  lines  w'  x' ,  y'  z',  2'—!',  and  -£'—>'•  Where  these 
horizontal  lines  intersect  the  curved  line  representing  the 
net  area  required  for  the  combined  uniform  and  concen- 
trated loads,  draw  short  vertical  lines  to  the  next  horizontal 
line  above;  draw  with  heavy  lines  the  rectangles  represent- 
ing" the  flange  plates  and  flange  angles;  scale  the  length  of 
the  flange  plates  with  the  scale  to  which  the  span  c  d  was 
laid  off,  and  the  theoretical  length  of  the  flange  plates  will 
be  found. 

In  the  girder  under  consideration,  the  theoretical  length 
of  the  top,  or  first,  plate  is  found  to  be  17  ft.  1  in. ;  the  length 
of  the  second  plate,  29  ft.  G  in. ;  the  length  of  the  third 
plate,  39  ft.  0  in. ;  and  the  length  of  the  last,  or  fourth,  plate 
is  46  ft.  0  in. 


90  ARCHITECTURAL  ENGINEERING.  §  6 

RIVET  SPACING. 

49.  Rivets  in  the  End  Angles  or  Stiffeiiers  Over  the 
Abutment. — First,  the  allowable  safe  load  upon  the  rivet 
should  be  determined.     Whether  the  double  shear  of  the 
rivet,  or  the  bearing  value  of  the  plate  around  the  rivet  hole, 
is  greatest,  should  be  found  as  previously  explained.    Having 
obtained  the  safe  allowable  load  for  each  rivet,  a  sufficient 
number  should  be  placed  in  the  end  angles  or  stiff eners  to 
take  care  of  the  entire  shear  at  that  point,  which  on  a  simple 
girder  is  equal  to  the  end  reaction  at  the  abutment.     For 
example,  let  the  end  reaction  of  a  girder  be  100,000  pounds; 
|-inch  rivets  are  used  and  the  thickness  of  the  web  is  f  inch. 
With  an  allowable  tensile  stress  of  12,000  pounds  per  square 
inch,  the  web-bearing  value  of  a  f-inch  plate  is  6,825  pounds, 
which  is  the  allowable  load  on  the  rivet.     The  number  of 
rivets  required  in  the  two  pairs  of  end  angles  is,  therefore, 
100,000-=-6,825  =  14.6,  say  16,  or  8  rivets  in  each  pair. 

50.  Rivets   in   the   Stiffeners   Between   the   Abut- 
ments.— If  possible  the  rivets  in  the  intermediate  stiffeners 
are  usually  spaced  the  same  as  in  the  end  stiffeners.     It  is 
hardly  possible  to  make  any  calculation  of  practical  value  in 
regard  to  the  number  and  spacing  of  these  rivets,  and  in 
fact  no  calculation  is  required ;  a  practical  rule  is  that  the 
pitch   of  these   rivets   should  never  exceed  6  inches,   nor 
should  it  exceed  16  times  the  thickness  of  the  leg  of  the  angle. 

51.  Rivets    Connecting    Flange    Angles   With    the 
Web. — It  will  readily  be  seen  that  when  a  plate  girder  is 
loaded,   the  tendency  of  the  flanges  and  angles  is  to  slide 
horizontally  past  the  web;   this  tendency  to  slide  induces 
a    horizontal    flange    stress.       The    rivets     connecting   the 
angles  to  the  web  resist  this  tendency,  and  there  must  be 
a  sufficient  number  of  rivets  to  do  it  safely. 

The  stress  which  at  any  point  is  transmitted  horizontally 
from  the  web  to  the  flange  is  equal  to  the  increment  of  the 
flange  stress  at  that  point.  This  increment  is  found  by 
dividing  the  maximum  shear  at  any  point  by  the  depth  of 
the  girder. 


6 


ARCHITECTURAL  ENGINEERING. 


1)1 


For  example,  assume  a  girder  of  40  feet  span,  as  shown 
in  Fig.  G4,  with  a  depth  of  4  feet,  and  a  uniformly  dis- 
tributed load  of  200,000  pounds.  The  shearing  stress 

in  the  girder  at  the  left  reac- 
tion, or  point  a,  is  equal 
to  A1?  in  this  case  100,000 
pounds.  At  b,  4  feet  from 
A',,  the  vertical  shear  in  the 
girder  is  100,000  -  (5,000  X  4) 
=  80,000  pounds;  at  c,  8  feet 
from  A5,,  the  vertical  shear  is 
100,000  -  (5,000  X  8)  =  GO, 000 
pounds;  at  d  it  is  100,000 
-(5,000X12)  =  40, 000  pounds; 
at  c  the  shear  is  20,000  pounds; 
and  at  f  the  .shear  is  zero. 

From  the  above  results,  the 
rate  of  increase,  per  inch  of 
length,  of  the  horizontal  stress 
in  the  flange  at  the  several 
points  a,  b,  c,  d,  c,  and  /  may  be 
obtained  by  dividing  the  shear 
at  those  points  by  the  depth 
of  the  girder  in  inches.  Thus, 
at  the  end  a  of  the  girder  the 
the  horizontal 


increase      in 


flange  stress  is 


100, 000 


=  2,083 


4X12 
pounds  per  inch  of  run;  at  /;, 


80,000 


=  1,6G7  pounds  per  inch 


of  run;    at  c, 


6°'°™  =  1,250 


4X12 
pounds  per  inch  of  run;  at//, 


40,000 


=  833  pounds  per  inch  of 


run ;  and  at  e, 


4x12 


4X12 
=  417  pounds  per  inch  of  run. 


92  ARCHITECTURAL  ENGINEERING.  §  6 

If  |--inch  rivets  are  used,  the  allowable  safe  load  for 
each  rivet,  based  on  a  fiber  stress  of  12,000  pounds  per 
square  inch,  with  web  bearing-  in  a  f-inch  plate,  is  6,825 
pounds.  Then  at  the  end,  where  the  increase  in  stress  is 
2,083  pounds  per  inch  of  run,  the  rivets  must  not  be  spaced 
farther  apart  than  6,825-^2,083  =  3.27  inches,  from  center 
to  center.  At  I?,  the  maximum  allowable  pitch  of  the 
rivets  is  6, 825 -i- 1,667  =  4.09  inches;  at  c,  the  pitch  may 
be  6,825-4-1,250  =  5.46,  or  about  5^,  inches;  and  at  d, 
6, 825  -T-  833  =  8. 19  inches.  Since,  for  practical  reasons,  the 
rivets  in  the  vertical  leg  of  the  flange  are  spaced  the  same 
in  both  the  upper  and  lower  chords,  and,  since  the  greatest 
allowable  pitch  of  rivets  in  a  compression  member  is 
6  inches,  it  is  evident  that  it  is  needless  to  carry  the 
calculation  further. 

In  accordance  with  the  above  calculation,  the  pitch  of  the 
rivets  between  a  and  b  should  be  3^  inches;  between  b  and  c 
the  pitch  should  be  about  4  inches;  while  between  c  and  d 
the  theoretical  pitch  is  5^  inches;  since  the  theoretical  pitch 
at  d  is  more  than  6  inches,  which,  for  practical  reasons,  is 
not  allowable,  all  the  rivets  between  d  and  the  center  of  the 
girder  should  be  spaced  6  inches  from  center  to  center. 

52.  Effect  of  Vertical  Stress. — Sometimes  the  vertical 
as  well  as  the  horizontal  stress  in  the  flange  is  taken  into 
account  in  spacing  the  rivets,  in  which  case,  the  resultant 
of  the  two  stresses  is  the  stress  that  must  be  provided  for. 
The  vertical  stress  is  due  directly  to  the  load  resting  upon 
the  flange  of  the  girder,  which,  through  the  rivets,  is  trans- 
mitted to  the  web-plate. 

In  the  plate  girder  shown  in  Fig.  64,  the  increase  in  the 
horizontal  flange  stress  at  the  end  is,  as  previously  calcu- 
lated, 2,083  pounds  per  inch  of  run;  the  load  upon  the 
girder  being  uniformly  distributed,  the  vertical  stress  on 
the  flange,  per  lineal  inch,  is  equal  to  the  entire  load  on  the 
girder  divided  by  the  span  of  the  girder  in  inches;  it  is, 
therefore,  200, 000 -r-  (40  X  12)  —  416  pounds  per  inch  of  run. 

The  total  stress  to  be  resisted  by  the  rivets  is,  therefore, 


§  G  ARCHITECTURAL  ENGINEERING.  03 

equal  to  the  resultant  of  2,(>83  pounds — due  to  the  increase 
in  the  horizontal  stress  on  the  flange — and  the  vertical  stress 
of  416  pounds;  this  resultant  is  4  2^3"  +  410*  =  2. 124 
pounds  per  inch  of  run.  The  pitch  of  the  rivets  at  the  end 
of  the  girder  would  then  be  0,825 -=-2, 124  =  3.21,  approxi- 
mately 3^,  inches. 

At  b,  4  feet  from  the  end  of  the  girder,  the  horizontal 
increment  of  stress  on  the  flange,  as  previously  calculated,  is 
1,607  pounds  per  inch  of  run,  while  the  vertical  stress 
remains  the  same;  the  combined  action  of  these  two  forces 
produces  a  resultant  stress  upon  the  rivets  of  \  1,007'  -j-410" 
=  1,717  pounds  per  inch  of  run,  and  this  divided  by  the 
value  of  one  rivet  gives  0,825 -h  1,71  7  —  3. !>7,  nearly  4, 
inches.  Similar  calculations  may  be  made  for  each  panel 
point  to  the  center  of  the  girder,  or  until  the  pitch  exceeds 
the  allowable  limit  of  6  inches. 

The  above  results  show  that  the  values  of  the  pitch  in 
which  the  vertical  stress  due  to  the  load  is  taken  into 
account,  are  nearly  the  same  as  those  first  obtained;  the 
effect  of  the  vertical  stress  has,  therefore,  little  influence  on 
the  pitch  of  the  rivets,  and  it  is  hardly  necessary  to  go 
into  such  refinement  in  the  design  of  an  ordinary  plate 
girder. 

53.  Rivets  Spaced  According  to  tlie  Stress  Pro- 
duced by  the  Bending  Moment.— The  rivets  which  con- 
nect the  flange  angles  with  the  web-plate  may  also  be  spaced 
according  to  the  stresses  produced  on  the  flanges  by  the 
bending  moment. 

The  horizontal  stress  on  the  flanges  diminishes  cither  way 
from  the  point  of  greatest  bending  moment  towards  the  end 
reactions,  where  it  becomes  zero,  and  for  any  point  this 
stress  may  be  calculated  by  the  application  of  the  principle 
of  moments. 

If  the  bending  moment  is  obtained  at  any  panel  point  and 
is  divided  by  the  depth  of  the  girder,  the  stress  on  the 
flange  at  that  point  will  be  obtained;  and,  if  this  stress  is 
divided  by  the  allowable  load  upon  one  rivet,  the  number  of 


94  ARCHITECTURAL  ENGINEERING.  §  6 

rivets  required  between  that  point  and  the  end  reaction  will 
be  obtained. 

For  example,  in  the  girder  used  in  the  previous  illustra- 
tion (Fig.  64),  the  span  being  40  feet  and  the  load  200,000 
pounds,  the  bending  moment  at  the  center  is  equal  to 

WL       200,000X40 

— —  =  -     —g—     -  =  1,000, 000 foot- pounds;  then  the  depth 

of  the  girder  being  4  feet,  the  flange  stress  at  this  point  is 
1,000, 000-s- 4  =  250,000  pounds.  The  allowable  load  upon 
each  rivet  being  6,825  pounds,  the  number  of  rivets  between 
the  center  and  the  end  reaction  is  250,000-7-6,825  =  37  rivets, 
approximately. 

Now,  although  the  number  of  rivets  between  the  end 
reaction  and  the  center  of  the  girder  has  been  obtained,  the 
pitch  of  these  rivets  is  still  unknown.  The  rate  at  which 
the  horizontal  stress  in  the  flange  increases  varies,  being 
greatest  at  the  ends  and  least  under  the  position  of  maxi- 
mum bending  moment,  it  follows  that  the  rivets  should 
be  spaced  nearer  together  at  the  ends,  with  an  increase 
in  the  spacing  towards  the  point  of  greatest  bending 
moment. 

In  practical  work  the  rivet  spacing  is  seldom  varied  in 
any  one  panel ;  if,  however,  the  flange  stress  is  obtained  at 
each  of  the  stiffeners  b,  c,  d,  e,  and  /,  Fig.  64,  the  number 
of  rivets  required  between  each  of  these  points  and  the  end 
reaction  may  be  obtained ;  by  finding  the  difference  between 
these  numbers  for  any  two  consecutive  stiffeners,  the  number 
of  rivets  required  in  the  panel  between  those  stiffeners  is 
arrived  at.  For  example,  the  stresses  on  the  flange  at  each 
of  the  stiffeners  of  the  girder  shown  in  Fig.  64  are  as 
follows : 

BENDING  MOMENT.     DEPTH  OF  GIRDER.     FLANGE  STRESS. 
FOOT-POUNDS.  FEET.  POUNDS. 

At  b,  360,000  -f-  4  90,000 

At  c,  640,000  -*-  4  160,000 

At  d,  840,000  -=-  4  210,000 

At  *,  960,000  -r-  4  240,000 

At/,  1,000,000  -5-  4  =          250,000 


§  6  ARCHITECTURAL  ENGINEERING.  1(5 

The  approximate  number  of  rivets  between  each  stiffener 
and  the  reaction  Rt  is  as  follows: 

Between  b  and  R^  90, 000  -4-  6,825  =  13  rivets. 
Between  c  and  R^  160,000-7-6,825  =  23  rivets. 
Between  d  and  R^  210, 000  -4-  6,825  =  31  rivets. 
Between  c  and  A\,  240, 000 -=-6, 825  =  35  rivets. 
Between  /  and  Rit  250,000-4-6,825  =  30  rivets. 

Then  the  number  of  rivets  that  are  required  is : 
Between  b  and  a,  13  —  0  =  13  rivets. 
Between  c  and  /?,  23  — 13  =  10  rivets. 
Between  d  and  c,  31  —  23  =  8  rivets. 
Between  c  and  d,  35  —  31  =  4  rivets. 
Between  f  and  c,  36  —  35  =  1  rivet. 

Consequently,  the  pitch  between  the  stiffeners  will  be  as 
follows : 

Between  b  and  a,  48-4-13  =  3.69  inches. 
Between  c  and  /;,  48-4-10  =  4. 80  inches. 
Between  d  and  c,  48-4-  7  =  6.85  inches. 

It  is  needless  to  continue  the  calculation  further,  because 
between  d  and  c  the  theoretical  pitch  exceeds  6  inches,  the 
limit  allowable  for  a  compression  member. 

By  the  first  method  the  pitch  at  each  stiffener  or  panel 
point  is  determined,  while  by  the  second  the  average 
pitch  between  two  consecutive  panel  points  is  obtained ; 
in  order  to  compare  the  two,  we  will  reduce  the  results 
obtained  by  the  first  to  the  basis  of  the  second.  In  the  first 
method  the  pitch  at  the  several  stiffeners  or  panel  points 
was  found  to  be : 

At  a  =  3.27  inches. 
At  b  —  4.09  inches. 
At  c  =  5.46  inches. 
At  d  =  8.19  inches. 

From  these  the  average  pitch  between  the  several  points 
would  be: 

Between  a  and  b,  (3.27  +  4.09)  -i- 2  =  3.68  inches. 
Between  b  and  c,  (4.09 +  5.46) -j-2  =  4.78  inches. 
Between  c  and  d,  (5.46  +  8. 19)  -4-  2  =  6.83  inches. 


90  ARCHITECTURAL  ENGINEERING.  §  6 

These,  on  comparison,  are  found  to  be  almost  identical 
with  the  values  3. 09,  4. 80,  and  0. 85  inches  obtained  by  the 
second  method. 

54.  Rivets  Spaced  According  to  the  Direct  Vertical 
Shear. — This  is  a  method  much  used  in  practical  work,  and 
will  be  found  to  give  safe  results,  corresponding  favorably 
with  those  obtained  by  the  previous  methods.     The  method 
is  based  on  the  assumption  that  at  any  point  the  horizontal 
shear  between  the  flange  angles  and  the  web-plate  is  equal 
to  the  vertical  shear  on  the  girder;  for  example,  the  vertical 
shear  at  the  end   stiffener  or  point  a,  Fig.  04,   is   100,000 
pounds ;  then,  according  to  this  method,  the  shearing  stress 
between   the    flange    angles   and    the  web-plate  is   100,000 
pounds,  distributed  over  the  space  between  the  panel  points 
a  and  b,  and  sufficient  rivets  should  be  placed  between  these 
points  to  safely  sustain  this  shear. 

The  allowable  web-bearing  load  on  a  ^-inch  rivet  in  a  f -inch 
plate  being  6,825  pounds,  the  number  of  rivets  required 
between  a  and  b  is  100,000-^-0,825  —  15,  approximately;  the 
vertical  shear  at  b  is  80,000  pounds,  and  80,000  -=-  6,825  =  12, 
approximately,  the  number  of  rivets  to  be  used  between  b 
and  c-  the  shear  at  c  is  60, 000  pounds,  and  60, 000  ^6, 825  =  9, 
approximately,  the  number  of  rivets  to  be  used  between  c 
and  d;  similarly,  the  number  of  rivets  required  between  d 
and  c  is  found  to  be  6.  According  to  these  results,  the  pitch 
of  the  rivets  between  a  and  b  should  be  48 -=-15  =  3.2 
inches;  between  /;  and  c,  48-^-12  =  4  inches;  between  c 
and  </,  5.33  inches;  and  from  there  on,  6  inches. 

Rivet  spacing  in  plate  girders  is  governed  so  largely  by 
practical  considerations,  that  this  method  is  to  be  recom- 
mended on  account  of  its  convenience.  It  gives  safe  results 
which  agree  closely  with  those  obtained  by  the  more  cum- 
bersome methods. 

55.  Rivet  Spacing  in  the  Flange  Plates. — In  spacing 
the  rivets  which  bind  the  several  flange  plates  together,  a 
sufficient  number  of  rivets,  spaced  from  2f  to  3  inches  on 


§  G  ARCHITECTURAL  ENGINEERING.  97 

centers,  should  be  used  at  the  ends  of  each  plate  to  transmit 
the  allowable  stress  in  it  to  the  members  below.  For  the 
remainder  of  the  plate,  the  rivets  should  have  the  greatest 
allowable  pitch  for  a  compression  member;  that  is,  10 
times  the  thickness  of  the  thinnest  outside  plate,  providing 
such  a  distance  docs  not  exceed  G  inches.  To  illustrate  : 

An  intermediate  flange  plate  in  a  certain  girder  is  %  in.x  12 
in.,  the  sectional  area  thus  being  4^  square  inches.  From 
this  area  is  to  be  deducted  the  section  cut  out  by  two  £-inch 
rivet  holes,  (Ixf)x2  =  |  square  inch;  then  the  net  area  of 
the  cover-plate  is  4£  —  f  =  3|  square  inches.  Assuming 
that  a  safe  fiber  stress  of  15,000  pounds  was  used  in  calcu- 
lating the  strength  of  the  girder,  the  safe  strength  of  the 
cover-plate  is  3fx  15,000  =  50,250  pounds.  Now  the  safe 
load  on  a  |-inch  rivet  depends,  in  this  position,  on  the  ordi- 
nary bearing  value  of  a  f-ineh  plate,  which,  calculated  on 
the  basis  of  a  fiber  stress  of  32,000  pounds,  is  5,1 1!)  pounds. 
Hence,  the  number  of  rivets  required  in  the  end  of  this 
cover-plate  is  56,2504-5,11!)  =  1  <).!»,  say  12;  this  gives  0  on 
each  side  of  the  web,  and  they  should  be  spaced  about  3 
inches  from  center  to  center.  The  remaining  rivets  in  this 
plate  may  have  the  greatest  allowable  pitch  until  the  next 
cover-plate  is  reached. 


PRACTICAL    PROBLEMS. 

5C.  In  order  to  illustrate  the  application  of  the  rules 
and  formulas  given  above,  the  following  practical  problem 
will  be  assumed  and  worked  out: 

The  floor  of  a  building  used  for  light  manufacturing  pur- 
poses is  to  be  supported  by  three  plate  girders,  as  shown  at 
a,  a,  a,  Fig.  G5.  The  floor  is  composed  of  1-inch  yellow- 
pine  flooring  laid  upon  3"xl2"  hemlock  joists,  spaced 
16-inch  centers;  these  joists  are  to  carry  a  plastered  ceiling 
on  the  under  side.  The  live  load  upon  the  floor  will  be  80 
pounds  per  square  foot.  The  girder  itself  is  to  extend  below 
the  surface  of  the  ceiling  and  is  to  be  painted,  a  detail  of  the 
construction  being  shown  in  Fig.  GO. 

1    28 


08 


ARCHITECTURAL  ENGINEERING. 


The  total  load  upon  each  square  foot  of  floor  surface  is 

as  follows: 

Live  load,  per  square  foot  of  floor  surface . .        80  pounds. 

Lath  and  plaster,  per  square  foot  of  floor  sur- 
face   8  pounds. 

1-inch  yellow-pine  flooring,  per  square  foot 

of  floor  surface 4  pounds. 

Hemlock  joist  flooring,  per  square  foot  of 

floor  surface 6  pounds. 

Girder  (assumed),  per  square  foot  of  floor 

surface 8  pounds. 


Total 106  pounds. 


Baar/rig>  B/oek. 

" 


*-3XIS  Hemlock 


FIG.  65. 


The  floor  area  to  be  supported  by  one  girder  is  60x17.5 
=  1,050  square  feet;  and  the  total  uniformly  distributed 
load  upon  the  girder  is  1,050x106  —  111,300  pounds. 


6 


ARCHITECTURAL  ENGINEERING. 


99 


The  greatest  bending  moment  on  the  girder  is 

WL        111,300X00 

-  =  834, 7oO  foot-pounds. 


M  = 


8 


8 


/  'YC//OW  Pine  F/oor/'ng. 


"If   "  " 

))     3X/S  Hem/oc/(Jo/'St.       } 

I         r\         i 

ii 

_J                                                   1 

Defai/ -Showing  Joist  and  Flooring. 


^^,^J>^^,^, 

^ 

o   o 

3 

aoooaaoao 

3 

O     0     /  . 

3 

J 

-> 

=> 

L_ 

3 

J 

-^ 

FIG.  66. 


The  depth  of  the  girder  is  4  feet,  and  the  allowable  unit 
fiber  stress  to-be  used  is  15,000  pounds;  therefore,  the 
required  flange  area  may  be  determined  by  formula  14, 
Art.  43.  Substituting  the  proper  values  in  the  formula, 


we  have 


A  = 


834,750 


=  13.91  square  inches. 


4X15,000 

Assume  a  flange,  composed  of  two  5"x5"xTV"  angles  and 
two  |"X12"  flange  plates;  a  sketch  of  the  section  with  the 
location  of  the  rivets  is  shown  in  Fig.  67.  The  entire  area 
of  the  flange  is 

2-fX  12"  plates           =    9         square  inches. 
2-5"  X  5"  X  TV'  angles  =    8.3  G  square  inches. 
Total,     .     .     .      1  7. 3  6  square  inches. 
The  sectional  areas  cut  out  for  rivet  holes  are : 
4-£-inch  holes  through  f-inch  plate    =1.312  square  inches. 
4-J-inch  holes  through  f^-inch  angles  =  1.5  3  1  square  inches. 
2.8  4  3  square  inches. 


100 


ARCHITECTURAL  ENGINEERING. 


6 


The   net   area   of   the    flange   is,    therefore,    17.36  —  2.84 
=  14.52  square  inches,   which,   since  the  required   area  is 

13.91    square    inches,    is    ample, 
and  this  section  will  be  adopted. 

We  will  now  determine  the 
thickness  of  the  web-plate.  The 
reaction  at  either  end  is  equal  to 
one-half  of  the  load,  or  55,650 
pounds.  Assuming  that  there 
are  eleven  |~inch  holes  cut  in  line 
through  the  web-plate,  the  net 
depth  of  the  plate  will  be  48-11X.875  =  38.375  inches. 
Using  an  allowable  unit  shearing  stress  of  11,000  pounds, 
the  theoretical  thickness  of  the  web-plate,  from  formula  12, 

Art.  38,  is 

55650 


38.375  X  11.000        - 

It  is  not,  however,  practicable  to  use  this  thickness  of 
metal  for  a  web-plate,  since  it  would  not  provide  sufficient 
bearing  value  for  the  rivets.  As  it  is  never  good  practice 
to  use  a  web-plate  less  than  y5^  inch  in  thickness,  this  size 
will  be  adopted.  • 

The  lengths  of  the  flange  plates  are  now  required  ;  they 
may  be  determined  either  by  the  graphical  method  or 
by  formula  15,  Art.  44;  using  the  latter  method,  the 
theoretical  length  of  the  outside  plate  is  found  to  be 

/  =  GO  A/'.—-  i  -  30.87  feet,  or  about  30  ft.  10  in.,  to  which 
r     14.52 

is  to  be  added  1  foot  at  each  end  to  allow  for  riveting.    The 

total  length  of  the  plate  is,  therefore,  32  ft.  10  in.,  say  33  feet. 

Applying  the  formula  again,  the  length   of  the  second 


flange  plate  is  /  = 


=  43.66  feet,  or  about  43  ft. 
14.52 


8  in.  ;  adding  a  foot  at  each  end  gives  us  45  ft.  8  in. 

Consider  now  the  size  of  the  four  stiff  eners  at  the  end 
of  the  girder.  The  reaction  at  the  end  of  the  girder  is 
55,650  pounds,  and  the  allowable  compressive  strength  of 


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§  G  ARCHITECTURAL  ENGINEERING.  lul 

the  material  in  the  girder  will  be  taken  at  13.000  pounds. 
Then  the  sectional  area  required  in  the  four  angles  compo- 
sing the  stiffeners  on  the  plate  girder  over  the  abutments  is 
55,050-4- 13,  < '00  =  4.28  square  inches.  Since  it  would  not 
be  advisable  to  use  smaller  than  a  4"  X  4"  X  TV  angle  in  this 
position,  the  sectional  area  of  which  is  2.4  square  inches  (see 
table  "  Properties  of  Angles  ").  it  is  evident  that  there  will  be 
ample  strength  in  the  four  stiffeners.  The  other  stiffeners 
may  be  made  of  3"x3"XyV  angles,  which  is  the  smallest 
size  that  should  be  used  for  any  girder  requiring  interme- 
diate stiffeners. 

The  rivet  spacing,  etc.  needs  no  explanation ;  it  would  be 
well,  however,  for  the  student  to  calculate  the  number  of 
rivets  for  the  several  parts  and  compare  the  results  with 
the  number  actually  used  as  shown  by  the  detail  drawing, 
Fig.  68.  He  will  undoubtedly  find  that  more  rivets  are  used 
than  are  actually  required,  but  he  must  bear  in  mind  that 
there  are  always  practical  considerations  which  influence 
more  or  less  the  design  of  structural  work. 

In  Fig.  68  it  will  be  noticed  that  the  web-plate  is  spliced 
at  the  point  a.  The  shear  at  this  point  is  equal  to  55,650 
pounds,  the  reaction  at  Rt  minus  the  load  on  the  girder 
between  R{  and  the  point  a  under  consideration,  that  is,  to 
55,650  —  37,150  =  18,500  pounds.  A  sufficient  number  of 
rivets  must  be  placed  on  the  two  sides  of  the  joint  to  take 
care  of  this  shear  safely. 

Fig.  69  shows  the  design  of  a  heavily  loaded  girder  with 
long  span ;  this  girder-  was  designed  to  carry-  a  uniformly 
distributed  load  of  2,400  pounds  per  lineal  foot  and  two  con- 
centrated loads  of  60,000  pounds,  placed  one  on  each  side  of 
the  center  of  the  girder  and  12  feet  6  inches  therefrom. 
The  unit  fiber  stress  allowable  in  calculating  the  flange  sec- 
tion was  taken  at  15,000  pounds. 

The  student  should  note  particularly  the  splices  on  this 
girder.  In  this  case  it  was  necessary  to  splice  the  flange 
angles;  when  this  is  done,  care  must  be  taken  to  weaken  the 
sectional  area  of  the  angle  as  little  as  possible  by  the  punching 
of  the  rivet  holes,  and  a  sufficient  number  of  rivets  must 


102  ARCHITECTURAL  ENGINEERING.  §  6 

also  be  placed  each  side  of  the  joint,  so  that  the  resistance  of 
the  rivet  section  may  equal  that  of  the  net  section  of  the 
flange  angles.  A  careful  study  should  be  made  of  the  vari- 
ous other  details  on  this  drawing,  which  represents  excellent 
modern  practice. 


DEFLECTION    OF    BEAMS. 

57.  Elasticity  is  that  property  which  a  body  possesses 
of  returning  to  its  original  form  after  being  strained  or  dis- 
torted by  the   application  of  a  stress.       This    property   is 
possessed  by  all  bodies  in  a  greater  or  less  degree.     If,  after 
being  distorted,  a  body  does  not  perfectly  resume  its  original 
form,  it  is  said  to  have  a  permanent  set. 

58.  Elasticity  of  Building  Materials.— It  is  believed 
that  the  elasticity  of  all  solids  is  more  or  less  imperfect,  and 
that  the  slightest  strain  produces  a  corresponding  permanent 
set.     It  is  customary,  however,  to  consider  the  elasticity  of 
all  building  material   as  practically  perfect  within  certain 
limits.     Under   this   assumption,  stresses,  up   to   a   certain 
limit,  may  be  applied  and  removed,  and  the  resulting  strain 
or  alteration  of  form  will  be  only  temporary,  with  no  appre- 
ciable permanent  set.       Stresses   above    this   limit   would, 
however,  cause  permanent  sets. 

59.  The  elastic  limit  of  any  material  is  the  maximum 
unit  stress  that  may  .be  applied  to  it  without  causing  any 
apparent  permanent  set. 

To  illustrate :  Consider  a  piece  of  steel  wire  supported  at 
one  end  and  loaded  by  a  weight  suspended  from  the  other. 
The  wire  is  found  to  stretch  under  the  action  of  the  load, 
and  by  varying  the  weight  or  stress,  the  strain  in  each  case 
is  found  to  vary  in  the  same  proportion,  so  long  as  the 
weight  is  not  greater  than  one-fourth  of  the  breaking 
strength  of  the  wire;  within  this  limit  it  is  found  upon 
removing  the  weight  that  the  wire  resumes  its  original 
length. 

If  the  load  is  made  considerably  greater  than  one-fourth  of 


§  6  ARCHITECTURAL  ENGINEERING.  103 

the  breaking-  strength  of  the  wire,  it  is  found  that  when 
the  load  is  removed  the  wire  has  taken  a  permanent  set;  in 
other  words,  it  will  not  return  to  its  original  length.  If  the 
wire  remains  permanently  longer  than  it  was  before  the 
load  was  applied,  it  has  been  strained  beyond  the  limit  of 
elasticity. 

Suppose  a  weight  of  2,000  pounds  is  hung  from  the 
end  of  a  wrought-iron  rod  having  a  sectional  area  of  1 
square  inch,  and  the  rod  stretches  about  T7^7  part  of  its 
original  length.  When  the  weight  is  removed,  the  bar 
resumes  its  original  length,  as  far  as  can  be  measured  by 
ordinary  instruments.  Now  instead  of  2,000  pounds,  attach 
a  weight  of  24,000  pounds  to  the  rod,  and  it  stretches  about 
TffW  Paft  of  its  length;  when  this  weight  is  removed,  we  find 
that  the  bar  does  not  return  to  its  original  length,  and  that  it 
is  slightly  longer  than  it  was  before;  that  is,  the  bar  has  a 
permanent  set. 

The  unit  stress,  where  the  weight  upon  the  rod  is  just 
sufficient  to  produce  the  least  permanent  set,  is  called  the 
elastic  limit. 

60.  The  modulus  of  elasticity  is  the  ratio  of  the  unit 
stress  to  the  unit  strain  for  loads  within  the  elastic  limit. 

For  example,  if  the  weight  of  2,000  pounds  upon  the  iron 
bar  whose  section  is  1  square  inch  produces  an  elongation 
of  -3-3^7  of  the  original  length  of  the  bar,  the  unit  stress  is 
2,000  pounds  per  square  inch;  the  unit  strain  is  y^^^;  and 
the  modulus  of  elasticity  is  2,000-=- 3-^^  =  20,000,000 
pounds  per  square  inch. 

In  most  building  materials  the  modulus  of  elasticity  for 
tension  and  the  modulus  for  compressive  stresses  may  be 
considered  as  practically  equal.  The  modulus  of  elasticity 
of  the  principal  building  materials  used  in  the  construction 
of  beams  are  given  in  the  table  "Modulus  of  Elasticity  of 
Metals." 

61.  Deflection  is  the  name  applied  to  the  distortion  or 
bending  produced  in  a  beam  when  subjected  to  transverse 
stresses.     If,  upon  the  removal  of  the  transverse  stresses  or 


104  ARCHITECTURAL  ENGINEERING.  §  G 

loads  upon  the  beam,  it  returns  to  the  straight  or  original 
form,  the  material  in  the  beam  has  not  been  strained  beyond 
the  elastic  limit.  On  the  other  hand,  if  the  internal  stresses 
exceed  the  elastic  limit  of  the  material,  a  permanent  set  will 
be  given  the  beam. 

(52.  Stiffness  is  a  measure  of  the  ability  of  a  body  to 
resist  bending;  this  property  is  very  different  from  the 
strength  of  the  material  or  its  power  to  resist  rupture. 

The  stiffness  of  a  structure  does  not  depend  so  much  upon 
the  elasticity  of  the  material  of  which  it  is  composed  as  upon 
its  arrangement  and  form ;  for  example,  a  floor  may  be  built 
of  shallow  and  wide  joists  which  will  be  sufficiently  strong 
to  carry  a  given  load,  but  it  will  not  be  nearly  so  stiff  as  a 
floor  of  equal  strength  built  of  narrow  and  deep  ones.  This 
property  of  stiffness  is  as  important  in  building  construction 
as  mere  strength,  and  the  two  should  be  considered  together; 
thus,  the  floor  joists  of  a  building  may  be  strong  enough  to 
resist  breaking,  but  so  shallow  as  to  lack  stiffness,  in  which 
case  the  floor  will  be  springy  and  vibrate  under  the  foot- 
steps of  people  walking  upon  it.  If  there  is  a  plastered 
ceiling  on  the  under  side  of  the  joists  of  such  a  floor,  the 
deflection  of  the  joists  may  cause  the  plaster  to  crack  and 
fall  into  the  room  below.  Where  stiffness  is  lacking  in  the 
rafters  of  a  roof,  they  will  be  liable  to  sag,  thereby  causing 
unsightly  hollows  in  the  surface  of  the  roof,  in  \vhich  mois- 
ture and  snow  may  lodge,  which  would  be  very  detrimental 
to  the  roof  covering. 

63.  Deflection  of  Beams. — From  the  foregoing  it  is 
evident  that  not  only  must  the  strength  of  the  beams  com- 
posing a  structure  be  calculated  to  withstand  rupture,  but 
the  beams  must  be  stiff  or  rigid  enough  to  resist  bending. 
It  is,  therefore,  important  to  be  able  to  calculate  the  deflec- 
tion of  any  beam  under  its  load,  and  if  found  excessive,  the 
size  of  the  beam  may  be  increased  and  the  deflection  reduced 
to  working  limits. 

The  amount  of  deflection  that  exists  in  beams  loaded  and 


§  6  ARCHITECTURAL  ENGINEERING.  105 

supported  in  different  ways  may  be  calculated  by  the  .for- 
mulas given  in  the  table  "  Deflection  of  Beams."  In  using1 
these  formulas,  all  the  loads  should  be  expressed  in  pounds 
and  the  length  in  inches.  The  modulus  of  elasticity  is 
denoted  by  E,  and  the  moment  of  inertia  of  the  section  by  /. 

EXAMPLE  1. — A  10-inch  steel  I  beam,  supported  at  the  ends,  must 
support  a  uniformly  distributed  load  of  10,000  pounds.  The  span  of 
the  beam  is  20  feet,  and  its  moment  of  inertia  is  178;  there  is  to  be  a 
plastered  ceiling  on  its  under  side,  the  allowable  deflection  of  which 
is  Jjj  inch  for  each  foot  of  span.  Will  the  deflection  of  the  beam  be 
excessive  ? 

SOLUTION. — The  formula  for  the  deflection  of  a  beam  of  this  character 

5  /  [ '  /  ;| 

from    the    table    "Deflection  of  Beams"  is  I)  =  ;,T,-,— , T-V.     From  the 

oo4  /:,  / 

table  "  Modulus  of  Elasticity  of  Metals,"  the  modulus  of  elasticity  of 
structural  steel  is  found  to  be  29,000,000,  and  this,  with  the  values  given 
in  the  example,  substituted  in  the  formula,  gives  us  the  deflection 

5  x  10.000  XJUO:1__ 
~  384  X  29,000,000  X  178 
about  i  of  an  inch. 

Since  the  allowable  deflection  for  each  foot  of  span  is  -£-a  of  an  inch, 
the  total  allowable  deflection  is  ^  of  20  =  §  of  an  inch.  This  is  twice 
as  much  as  the  calculated  deflection,  and  the  beam  therefore  satisfies 
the  required  conditions.  Ans. 

EXAMPLE  2. — A  12"  X  16"  Northern  short-leaf,  yellow-pine  girder  must 
support  a  symmetrically  placed  triangular  piece  of  brickwork,  which 
weighs  about  12,000  pounds.  What  will  be  the  deflection  of  the  timber 
if  the  span  is  20  feet  ? 

SOLUTION. — The   formula   for  the  deflection  in  this  case  from  the 

I V  1 3 

table  "  Deflection  of  Beams,"  is  D  =  -^n   .,"/.     From  the  table  "  Modu- 

60  h  / 

lus  of  Elasticity  of  Timber,"  the  value  of  the  modulus  of  elasticity  is 
found  to  be  1,200,000.  The  moment  of  inertia  of  the  section,  from  the 
bd*  . 


formula  /  = 


L2 


Then,  by  substituting  the  given  values,  the  deflection  is 

12,000  X  2403 


D    =    ~,rf 


60X1,200,000X4,096 


about  A  of  an  inch.     Ans. 


n 


106  ARCHITECTURAL  ENGINEERING.  §  6 

EXAMPLES   FOR   PRACTICE. 

1.  The  moment  of  inertia  of  a  12-inch  steel  I  beam  is  220,  and  its 
span  is  25  feet.     If  the  ends  of  the  beam  are  simply  supported,  what 
will  be  its  deflection  under  a  concentrated  load  of  10,000  pounds  sus- 
pended from  its  center  ?  Ans.  .88  in. 

2.  A  cantilever  beam  of  12"  X  16"  Georgia  yellow  pine  extends  from 
a  building  wall  10  feet,  and  is  loaded  on  the  end  with  a  concentrated 
load  of  18,000  pounds.     What  will  be  the  greatest  deflection  of  the 
beam  ?  Ans.  1.49  in. 

3.  The  span  of  a  15-inch  steel  I  beam  is  30  feet,  and  the  moment  of 
inertia  of  its  section  is  429.6.     The  load  upon  the  beam  is  uniformly 
distributed  and  amounts  to  3,000  pounds  per  lineal  foot;  if  the  ends  of 
the  beam  are  firmly  fixed,  what  will  be  its  deflection  ?  Ans.  .88  in. 


FI.ITCH-PI.ATE    GIRDERS. 

64.  The  name  flitch-plate  girder  is  applied  to  a  beam 
composed  of  two  timbers  bolted  together  with  a  flat  plate  of 
iron  or  steel  sandwiched  between  them ;  hence  it  is  often 
called  a  sandwiched  girder. 

This  girder  is  used  where  it  is  desired  to  impose  a  greater 
weight  than  the  wooden  beams  will  support,  and  is  consid- 
ered somewhat  cheaper  than  a  steel  I  beam ;  owing,  how- 
ever, to  the  present  exceedingly  low  price  of  structural  steel, 
it  is  doubtful  if  there  is  now  much  difference  in  cost  in  favor 
of  the  flitch-plate  girder. 

The  great  advantage  a  beam  of  this  character  possesses 
over  a  steel  rolled  section  probably  lies  in  the  fact  that  it  is 
partially  fireproof;  the  wood  may  be  charred  considerably 
without  losing  all  its  strength,  and  it  will  protect  the  iron  or 
steel  plate  from  the  intensity  of  the  heat,  thus  acting  as  a 
non-conductor  and  fireproofing  material  for  the  metal. 

It  is  difficult  to  so  proportion  a  girder  of  this  character, 
as  to  be  economical  in  the  use  of  both  the  wood  and  the 
metal.  Theoretically,  the  iron  or  steel  plate  should  be  so 
proportioned  that,  when  carrying  its  share  of  the  load,  it 
deflects  equally  with  the  wooden  beams,  otherwise  there  is 
a  tendency  for  the  bolts  either  to  shear  off  or  to  crush  and 
tear  the  wood. 


§  G  ARCHITECTURAL  ENGINEERING.  107 

The  two  timbers  should  be  considered  as  a  single  beam, 
and  the  calculation  made  for  the  safe  load  they  will  sus- 
tain. The  plate  will  then  be  proportioned  to  support  the 
remainder  of  the  load  coming  upon  the  girder.  The  timber 
beams  and  the  plate  should  be  so  proportioned  that  they 
will  both  deflect  equally  under  their  respective  loads.  It  is, 
however,  almost  impossible  to  realize  such  a  condition  as 
this,  as  may  be  seen  by  referring  to  the  following  illustrative 
example : 

65.  The  span  of  a  flitch-plate  girder  is  20  feet,  and  it  is 
composed  of  two  yellow-pine  timbers,  each  G  in.  x  10  in., 
with  a  plate  of  steel  16  inches  wide  bolted  between  them.  The 
girder  carries  a  load  of  20,000  pounds  concentrated  at  the 
center,  and  is  to  be  so  proportioned  that  its  deflection  under 
this  load  will  not  be  likely  to  cause  plaster  cracks.  What 
should  be  the  thickness  of  the  plate  ? 

It  will  first  be  necessary  to  calculate  the  strength  of  the 
two  yellow-pine  timbers;  in  order  to  do  this,  the  section 
modulus  of  the  combined  cross-section  of  the  two  timbers 

/      V2 

will  be  obtained  by  the  formula  K  =  — — ,  which,  upon  sub- 

jr  12X16' 

stitutmg  the  values,  gives  us  A   =  — =  o!2. 

Now  the  safe  uniformly  distributed  load  W  that  a  rect- 
angular beam  will  support,  may  be  obtained  by  the  formula 

W=  ^?.  (17.) 

o  1^ 

This  formula  is  readily  derived  by  combining  formula  4 
with  formula  13,  Architectural  Engineering,  §  5.  Note, 
however,  that  in  formula  4,  the  bending  moment  is  in  incJi- 
pounds,  while  in  formula  13,  Architectural  Engineering,  §5, 
it  is  expressed  in  foot-pounds. 

It  is  known  that  a  beam  will  support  only  one-half  of 
this  load,  if  concentrated  at  the  center;  therefore,  the  for- 
mula for  obtaining  the  safe  concentrated  load  W  &t  the  center 
of  a  beam  is 


108  ARCHITECTURAL  ENGINEERING. 

2  K  S  KS 

W  =  mrX*  --"  3T> 
where  K  =  section  modulus  of  the  cross- section; 

5  =  safe  transverse  strength  of  the  material; 
L  =  span  of  the  beam  in  feet. 

For  the  beam  under  consideration,  K  was  found  to  be 
512.  5  may  be  obtained  by  dividing  7,300,  the  modulus  of 
rupture  for  yellow  pine,  as  given  in  Table  6,  Art.  61, 
Architectural  Engineering,  §5,  by  the  factor  of  safety, 
which  will  be  taken  as  5;  then,  5  =  7, 300 -J- 5  =  1,460 
pounds  per  square  inch ;  L  is  20  feet. 

With  the  above  values  substituted  in  the  formula,  the 
safe  load  that  may  be  carried  by  the  two  pine  beams  is  found 

„,       512x1,460  0. 

to  be  W  =  -  — •  =  12,4o8  pounds.     Since  the  entire 

3X20 

load  upon  the  girder  is  20,000  pounds,  the  load  which  the 
steel  plate  must  support  is  20,000  —  12,458  =  7,542  pounds. 

The  next  step  is  to  calculate  the  deflection  of  the  yellow- 
pine  timbers  under  their  load. 

The  deflection  of  a  beam  supported  at  both  ends  and 
loaded  with  a  concentrated  load  applied  at  the  center,  may 

WL* 

be  calculated  by  the  formula  D  =    .n  ,,  T\  see  table  "Deflec- 

48^7 

tion  of  Beams."  The  value  of  E  for  yellow  pine  will  be 
taken  at  1,200,000;  the  moment  of  inertia  of  the  section 

1 2  v  1  fi3 

is  7  =  -  -  =  4,096;    in    this    case    the    length  of    the 

12 

span  is  to  be  in  inches,  therefore,  L  =  20  X  12  =  240  inches; 
and  W,  according  to  the  problem,  is  12,458  pounds.  The 
substitution  of  these  values  in  the  formula  gives  us 
D  -  12,458  X2403 

:  48X1,200,000X4,096  = 

The  steel  plate  should  now  be  proportioned  so  that  it  will 
deflect,  under  its  load  of  7,542  pounds,  a  distance  equal  to 
.  73  inch,  the  deflection  of  the  wooden  girders. 

WL? 

The  formula  D  =  ,„  „  ,  may  be  transposed  and  written 
48  L  I       J 


§  G  ARCHITECTURAL  ENGINEERING.  10!) 

W 'L3 
I  —  -:  ,  ,-  /y     By  substituting  the  known  values  in  this  for- 

•±O  11.   1J 

T  7,542X2403 

mula,  we  have  /  =  -  —  102.  G 

48  X  29, 000, 000  X.  73 

Now  the  depth  of  the  plate  is  1G  inches,  and  /for  the 
required  section  is  102.G.  The  formula  /  =  — (—  may  be 

/X  12 
transposed  to  b  —  — j^~,   and  by  substituting  the  known 

102.GX  12 
values  in    the  latter,   it  becomes  (>  = =  .3mch, 

nearly,  the  required  thickness  of  the  plate. 

Before  finally  adopting-  this  thickness  for  the  plate,  how- 
ever, it  should  be  examined  to  determine  whether  the  deflec- 
tion of  .73  inch  causes  too  great  a  fiber  stress  on  the  steel. 
In  order  to  determine  the  fiber  stress  upon  the  plate,  the 
bending  moment  must  be  calculated.  According  to  Table  10, 
Art.  97,  Architectural  Engineering,  go,  the  formula  fora 

beam  with  a  concentrated   load  at  the  center  is  J/  =          "; 

7  542  X  20 
upon    substituting    the    values,   we    have    J/  =  - 

=  37,710  foot-pounds,  which,  multiplied  by  12,  gives 
452,520  inch-pounds.  The  section  modulus  of  the  section  is 

'}  v  1  fi2 

K  =  —     —  -  12.8. 
G 

Having  now  the  bending  moment  J/  and  the  section 
modulus  K,  the  unit  of  fiber  stress  .V  may  be  determined  by 

M 

the  formula  S"  =  -^?;  upon  substituting  the  values,  we  have 

„        452,520  . 

.S  =         '   ,  •   =  35,353   pounds    per  square    inch.       1  his   is 
12.  h 

entirely  too  great  and  should  be  reduced  to  about  one-third, 
or  say  12,000  pounds  per  square  inch.  In  order  to  obtain  this 
lower  unit  fiber  stress,  the  plate  must  be  about  three  times 
as  thick4  or  nearly  .9  inch;  a  -£-inch  plate  will  be  strong 
enough.  Changing  the  thickness  of  the  plate  in  this  manner 
reduces  the  deflection  of  both  the  timbers  and  the  steel 
plate  below  the  .73-inch  previously  determined,  and  the  full 


110 


ARCHITECTURAL  ENGINEERING. 


6 


strength  of  the  timbers  will  not  be  realized.  The  reduced 
deflection  will  be  an  advantage  in  this  case,  because 
.  73  inch  is  excessive.  The  deflection  should  not  be  more 
than  -^  of  an  inch  for  each  foot  of  span,  or  f  of  an  inch  in 
all.  Hence,  upon  increasing  the  thickness  of  the  plate,  and 
thus  decreasing  the  deflection,  the  deflection  of  the  girder 
will  probably  be  brought  within  the  desired  limits. 

The  bolts  holding  the  timbers  and  the  steel  plate  together 
are  best  located  along  the  neutral  axis ;  for  at  this  point 
there  would  practically  be  no  stress  upon  them.  But  to 
space  them  sufficiently  near  together  along  this  line  would 
tend  to  weaken  the  timber  too  much,  and  would  be  likely  to 
cause  the  destruction  of  the  beam  from  longitudinal  shear- 
ing along  this  line.  .  Therefore,  it  is  best  to  alternate  them 
above  and  below  the  line  of  the  neutral  axis,  thus  forming 


of  Neutral  Axis 


FIG.  TO. 

two  rows  of  bolts  as  shown  in  Fig.  70.  The  end  bolts  are 
doubled,  and  the  horizontal  distance  d  between  the  bolts  a 
and  b  should  be  about  equal  to  the  depth  of  the  girder.  The 
bolts  in  a  girder  of  this  character  should  be  about  1  inch  in 
diameter. 


KOOF  TRUSSES. 


DETERMINATION    OF    STRESSES    IN   THE 
FINK  TRUSS. 

66.  The  Polonceau  or  Fink  Truss,  shown  in  Fig.  71 
at  («),  is  a  favorite  form  of  truss.  It  is  often  built  of  rolled- 
.steel  sections;  it  is  also  built  with  wooden  rafter  members, 
structural-steel  struts,  and  wrought-iron  tension  members. 
Frequently  the  lower  chord  of  the  truss  is  cambered  (raised 


6 


ARCHITECTURAL  ENGINEERING. 


Ill 


at  the  center),  as  shown  at  (b),  Fig.  71.    Cambering-  the  lower 
chord  in  this  manner  gives   greater  headroom    under  the 


truss  at  the  center,  and  somewhat  improves  its  appearance ; 
but  it  increases  the  stresses  on  all  the  members  except  A'/, 
ML,  and  O  N. 

67.  Diagram  for  Vertical  Loads. — Fig.  72  is  a  frame 
diagram  showing  the  vertical  loads  upon  a  Fink  truss,  the 
span  of  which  is  80  feet,  the  lower  chord,  or  tie,  of  the  truss 
being  cambered  from  the  horizontal  28  inches.  The  stress 
diagram  for  the  vertical  loads  may  be  drawn,  as  shown  in 
Fig.  73,  by  first  drawing  the  vertical  load  line  af,  and  laying 
off  to  some  convenient  scale  the  loads  a  b,  b  c,  c  d,  d c,  and  cf, 
designated  in  the  frame  diagram,  Fig.  72,  by  A  B,  B  C,  CD, 
D  E,  and  E  F.  Since  the  truss  is  symmetrically  loaded, 
only  one-half  of  the  stress  diagram  need  be  drawn,  and 
consequently  the  loads  only  as  far  as  cf  need  be  laid  off  on 
the  vertical  load  line.  The  reactions  Rt  and  R^  are  each 
equal  to  one-half  of  the  load,  or  18,000  pounds;  hence,  the 
point  z  is  located  midway  between  c  and/",  and  za  represents 
the  reaction  R^  18,000  pounds. 


112 


ARCHITECTURAL  ENGINEERING. 


6 


The  stresses  around  the  joint  A  B  K  Z  may  be  drawn  in 
the  stress  diagram  by  commencing  at  b  and   drawing  b  k 

parallel  with  B  K, 

'}-o^-          H  an(j  from  z  a  ]me 

ooosr =ey  „  _ 

parallel  with  A  Z, 

intersecting  the 
first  line  at  k.  The 
polygon  of  forces 
around  this  joint 
will  then  be :  from 
a  to  b,  from  b  to  k, 
from  k  to  s,  and 
from  z  back  again 
to  a,  the  starting 
point.  From  the 
direction  of  these 
forces,  the  correct 
direction  of  the 
arrowheads  in  the 
frame  diagram 
may  be  marked, 
and  from  their 
direction  the  kind 
of  stress  upon 
the  member  is 
observed. 

The  next  joint 
in  the  truss  to  an- 
alyze is  B  C  L  K. 
In  the  stress  dia- 
gram begin  at  c 
and  draw  cl  par- 
allel with  C  L  in 
the  frame  diagram, 
then  from  /',  draw 
a  line  parallel  with 

L  K  in  the  frame  diagram,  until  it  intersects  the  line  drawn 
from  c  at  the  point  /.      The  polygon  of  forces  around  the 


ARCHITECTURAL  ENGINEERING. 


113 


joint  is  from  b  to  c ,  from  c  to  /,  from  /  to  k,  and  from  k 
back  to  d,  the  starting  point. 

Around  the  joint  K  L  ^[ Z,  the  stresses  are  obtained  by 
drawing   from    /  a    line    parallel  with    L  M  in  the    frame 


FIG.  73. 

diagram,  until  it  intersects  the  line  zk  at  the  point  m. 
The  polygon  of  forces  around  this  joint  is:  from  k  to  /, 
from  /  to  m,  from  in  to  2,  and  from  rr  back  again  to  /',  the 
starting  point. 

Difficulty  will  be  encountered  upon  attempting  to 
analyze  either  the  joint  C  DON  ML  or  M  N  Q  Z, 
for  at  each  of  these  joints  there  are  three  unknown 
forces  or  stresses.  It  is  impossible,  by  the  graphical 
method,  to  solve  the  stresses  around  a  joint  in  a  struc- 
ture where  there  are  more  than  two  unknown  forces, 
consequently  some  other  method  of  solution  must  be 
found. 

Upon  inspecting  the  frame  diagram,  Fig.  72,  it  will 
be  seen  that  the  joint  D  E  is  similar  to  the  joint  B  C\ 

1-29 


114  ARCHITECTURAL  ENGINEERING.  §  6 

the  panel  load,  4,500  pounds  in  each  case,  is  sup- 
ported by  two  forces,  B  K  and  K  L,  DO  and  O  P, 
respectively.  Since  the  directions  of  the  forces  are 
respectively  parallel,  and  the  panel  loads  are  equal,  the 
stresses  in  K  L  and  O  P  are  equal,  these  stresses 
being  due  only  to  a  component  of  their  respective  panel 
loads.  In  a  similar  manner,  it  can  be  shown  that  the 
stresses  in  K  L  and  OP  are  each  held  in  equilibrium 
by  the  pairs  of  forces,  K  Z  and  L  M,  ON  and  P  Q, 
and  that  the  stresses  in  L  M  and  N  O  are  produced 
by  equal  components  of  the  equal  stresses  in  K  L  and 
O  P ;  therefore,  the  stresses  in  L  M  and  N  O  are 
equal.  This  gives  us  the  magnitude  of  the  stress 
O  N,  its  equal  L  M  having  already  been  determined, 
and  the  diagram  can  be  completed  as  follows:  Draw 
;««  parallel  to  M  N  of  the  frame  diagram,  then  draw 
do  parallel  with  D  O,  and  of  such  a  length  that  when 
on  is  drawn,  from  its  extremity  o,  the  point  n  will  meet 
the  line  m  n  midway  between  the  lines  do  and  ep. 
This  construction  makes  the  length  on  equal  to  /;//, 
which  is  in  accordance  with  the  condition  that  the  stresses 
in  LM  and  O  TV  are  equal.  Then  the  polygon  of  forces 
around  this  joint  will  be:  from  c  to  d,  from  d  to  a,  from 
o  to  n,  from  n  to  m,  from  m  to  /,  and  from  /  to  c,  the 
starting  point. 

The  remaining  joints,  when  taken  in  their  usual  order, 
offer  no  difficulty,  and  the  other  half  of  the  diagram  need 
not  be  drawn  unless  it  is  desired  to  check  the  half  of  the 
diagram  just  completed. 

The  polygon  of  forces  that  has  just  been  traced  around 
the  joint  CDONML  affords  a  good  illustration  of  the 
rule,  that  the  forces  that  meet  at  a  joint  must  make  a  closed 
polygon  in  the  stress  diagram. 

One  of  the  peculiarities  of  the  stress  diagram,  Fig.  73,  and 
one  that  is  worthy  of  note,  as  it  will  materially  assist  in 
drawing  the  diagram,  is  that  the  triangles  I  km  and 
pon  are  equal  and  similar  to  the  larger  one  whose  base 
is  inn. 


6 


ARCHITECTURAL  ENGINEERING. 


115 


68.     The  wind-stress  diagram    may  now   be   drawn. 
First  the  frame  diagram  is  redrawn,  and  upon  it,  as  shown  in 


Fig.  74,  are  designated  the  wind  loads  acting  at  the  several 


116  ARCHITECTURAL  ENGINEERING.  §  6 

joints  of  the  truss,  in  a  direction  normal  to  the  slope  of 
the  roof. 

The  reactions  Rl  and  R3  may  be  calculated  by  the  princi- 
ple of  moments.  Since  the  truss  is  securely  fastened  at 
both  ends,  neither  end  being  free  to  move  in  a  lateral 
direction,  these  reactions  will  be  parallel  with  the  action  of 
the  wind  on  the  roof,  that  is,  normal  to  the  slope.  The 
reaction  Rt  may  first  be  obtained  by  extending  its  line  of 
direction  until  it  intersects  the  extension  of  the  left-hand 
rafter  member  at  the  point  a'.  Then  by  taking  the  center 
of  moments  at  the  left-hand  reaction  R^  the  magnitude  of 
the  reaction  Ry  may  be  computed. 

For  convenience,  reduce  to  feet  and  decimals  the  distance 
from  each  panel  point  to  the  point  of  rotation  ./?,;  the 
moments  about  this  point  will  then  be  as  follows : 

5,625  X  11.188  =  6  2,9  3  2.5  0  foot-pounds. 
5,625X22.375  =  12  5, 85  9. 38  foot-pounds. 
5,625  X  33.563  =  1  8  8,7  9  1.8  8  foot-pounds. 
2,813  X  44.750  =  1  2  5,8  8  1.7  5  foot-pounds. 
Total,  .  .  5  0  3,4  6  5.5  1  foot-pounds. 


The  distance  of  the  center  of  moments  of  the  line  of 
action  of  the  reaction  R^  is  44.7571-27  =  71.75  feet,  and 
503,465.51^-71.75  =  7,016,  the  magnitude  of  the  reaction 
R9,  in  pounds. 

Since  the  sum  of  the  wind  loads  is  22,501  pounds,  which 
is  equal  to  the  sum  of  the  reactions,  the  reaction  due  to  the 
wind  at  Rt  is  22,501  —  7,016  =  15,485  pounds. 

The  wind-stress  diagram,  Fig.  75,  may  now  be  drawn. 
First  draw  the  load  line  a  to  /  parallel  to  the  reactions  and 
direction  of  the  wind  loads  at  the  several  joints.  Then  lay 
off  on  the  load  line  the  loads  ad,  be,  c d,  de,  and  ef,  which 
are  respectively  equal  to  the  corresponding  loads  A  B,  B  C, 
CD,  DE,  and  E  Fin  the  frame  diagram.  Having  located 
the  point  f,  the  magnitude  of  the  reaction  R^  represented 
by  fzt  may  be  laid  off  upwards  (the  direction  in  which 


ARCHITECTURAL  ENGINEERING. 


117 


it  acts),  and  the  point  z  is  thus  located;  then  upon  sca- 
ling 5  a,  it  should  be  found  equal  to  15,485  pounds,  the 
reaction  A',. 

The  polygon  of  external  forces  will  then  be:  from  a  to  /;, 


Scale  '  to  J-JOO  Ib. 

Q 


FIG.  75. 

from  b  to  c,  from  c  to  d,  from  d  to  c,  from  c  to  f,  from  /  to  z, 
and  from  s  to  a,  the  starting  point. 

The  joint  CD  O  NML  may  be  solved  in  this  stress  dia- 
gram in  the  same  manner  as  it  was  solved  in  the  vertical 
load  stress  diagram. 

The  analysis  of  the  stresses  around  the  last  joint 
EFRQPin  the  frame  diagram,  is  interesting,  from  the 
fact  that  the  stress  q  r  closes  the  diagram,  and,  if  the  dia- 
gram is  correctly  drawn,  it  must  be  parallel  to  the  member 
QR. 

Having  drawn  both  the  wind  and  the  vertical  load  stress 
diagrams,  the  stresses  in  the  several  members  in  the  truss 
may  be  obtained  by  scaling,  and  their  magnitudes  may  be 
tabulated  as  follows: 


118 


ARCHITECTURAL  ENGINEERING. 


Member. 

Stress  Due  to 
Vertical  or  Dead 
Load. 

Stress  Due  to 
Wind  Load. 

Total  Stress. 

BK 

+  46,000 

+  34,500 

+  80,500 

CL 

+  44,000 

+  34,500 

+  78,500 

DO 

+  42,000 

+  34,500 

+  76,500 

EP 

+  40,000 

+  34,500 

+  74,500 

KL 

+   4,000 

+   5,500 

+   9,500 

MN 

+   8,000 

+  11,250 

+  19,250 

OP 

+   4,000 

+   5,500 

+   9,500 

ZK 

-41,500 

-36,750 

-  78,250 

ZM 

-35,500 

-28,500 

-64,000 

ZQ 

-21,000 

-11,000 

-32,000 

QN 

-15,500 

-  18,500 

-34,000 

QP 

-21,500 

-26,000 

-47,500 

FR 

+  17,000 

RQ 

-    2,000 

NO 

-    5,750 

-    8,000 

-13,750 

LM 

-    5,750 

-    8,000 

-13,750 

In  the  above  table,  compression  is  indicated  by  the  plus 
sign  and  tension  by  the  minus  sign. 


THE    DESIGN   OF    A    COMPOSITE    PIN-CONNECTED 
ROOF  TRUSS. 

69.  One  of  the  most  generally  used  forms  for  a  com- 
posite pin-connected  truss  is  the  Polonceau  or  Fink  truss, 
previously  described.  A  truss  of  this  pattern  is  usually 
made  of  wooden  rafter  members,  structural- steel  struts,  and 
wrought-iron  tension  members.  Where  the  wooden  rafter 
member  is  oiled,  or  otherwise  finished,  and  where  the  iron 
and  steel  members  are  tastily  painted,  these  trusses  make  a 
good  appearance  from  the  interior  of  the  room  or  building 
which  they  span,  and  are  adapted  for  supporting  the 


§  6  ARCHITECTURAL  ENGINEERING.  119 

covering  or  roof  for  such  buildings  as  mess  halls  of  bar- 
racks  or  asylums,  armory  drill  halls,  railroad  stations,  etc. 

In  order  to  illustrate  the  method  of  designing  the  mem- 
bers and  joints  in  a  truss  of  this  character,  we  will  now 
consider  the  design  of  the  truss,  Fig.  72,  the  stresses  in  which 
were  determined  by  the  diagrams  Figs.  73  and  75,  and  tabu- 
lated in  Art.  68. 

7O.  The  rafter  member  has  the  greatest  compressive 
stress  upon  it  between  the  points  A  B  and  B  C,  Fig.  72 ;  this 
stress  is  represented  in  the  table  of  stresses  by  the  stress  of 
80,500  pounds  in  the  member  B  K. 

In  addition  to  this  compressive  stress,  there  is  a  bending 
stress  in  the  rafter  due  to  the  construction  of  the  roof,  which 
is  composed  of  sheathing  laid  upon  joists  spaced  about  14 
inches  center  to  center  along  each  rafter.  The  wind  and 
vertical  loads  are,  by  this  construction,  transmitted  to  the 
panel  points  by  the  transverse  strength  of  the  rafter  between 
these  points;  it  will  therefore  be  necessary,  before  propqr- 
tioning  the  rafter  member,  to  calculate  the  stresses  due  to  the 
bending  moment  produced  by  the  dead  and  wind  loads. 

The  wind  load  acts  normally  to  the  rafter,  but  the  dead 
or  vertical  load  does  not.  If  great  accuracy  were  required, 
it  might  be  advisable  to  resolve  the  vertical  load  so  as  to 
determine  its  component  normal  to  the  roof,  and  add  this 
amount  to  the  wind  load  to  obtain  the  entire  load  normal  to 
the  rafter ;  such  refinement,  however,  will  not  be  necessary 
here,  and  the  vertical  load  and  wind  load  will  be  added 
together  directly  and  considered  as  a  uniformly  distributed 
load  upon  the  rafter  member  between  the  panel  points. 

The  sum  of  the  wind  and  dead  loads  for  a  panel  is  5,625 
+  4,500  =  10,125  pounds;  therefore  the  bending,  moment, 

,.        W L   .     ,,       10,125X11.188 
according  to  the  formula  M  =  -  — ,  is  M  —  - 

o  o 

=  14, 160 foot-pounds,  or  14,160x12  =  169, 920 inch-pounds. 
Assuming  that  the  rafter  member  is  made  of  yellow  pine, 
it  will  be  seen  by  referring  to  Table  6,  Art.  61,  Architec- 
tural Engineering,  §  5,  that  the  modulus  of  rupture  is  7,300; 
hence,  if  a  factor  of  safety  of  5  is  adopted,  the  safe  working 


120  ARCHITECTURAL  ENGINEERING.  §  6 

transverse  stress  in  the  material  will  be  7,  300  -=-5  =  1,460 
pounds.  Since  the  required  section  modulus  K  may  be 
obtained  by  dividing  M  (the  bending  moment  in  inch- 
pounds)  by  the  safe  unit  stress  S  of  the  material,  as 

M  „       169,920 

expressed  by  the  formula  K  -     -^,  we  have  K  — 

=  116,  the  section  modulus  required  to  resist  the  transverse 
stress  upon  the  rafter  member. 

Assume  that  a  depth  of  14  inches  is  adopted  for  the  rafter 
member;  then  by  transposing  formula  15,  Art.  1O1,  Archi- 


is  ,  ,,     . 

tectural  Engineering^  §  5,  K  =  -^—  to  b  =       ,9  ,  the  breadth, 

or  width,  of  the  rafter  member  required  to  resist  the  trans- 
verse stress,  by  substituting  the  values  of  K  and  d,  is 

"11^*  \/  R 

b  =  —  -75  —  =  3^-  inches.     Thus  it   is  seen  that  a  section 

of  3^-  in.  X  14  in.  yellow  pine  is  sufficient  to  take  care  of  the 
transverse  stress  upon  the  rafter  member.  But  in  addition  to 
this  transverse  stress  there  is  a  direct  compressive  stress  of 
80,500  pounds,  as  shown  by  the  stress  diagram,  which  must 
be  provided  for  in  the  same  member,  by  adding  material  in 
the  direction  of  its  width. 

From  Table  6,  Art.  61,  Architectural  Engineering,  §5, 
the  ultimate  compressive  strength  of  yellow  pine,  parallel 
to  the  grain,  is  found  to  be  4,  400  pounds  per  square  inch, 
and  as  a  factor  of  safety  of  5  is  used,  the  safe  compressive 
strength  will  be  4,  400  -r-  5  =  880  pounds  per  square  inch. 

Since  the  distance  from  joint  to  joint  is  not  great,  being 
only  about  11  feet,  the  rafter  member  between  the  joints 
need  not  be  considered  as  a  column,  but  it  may  be  designed 
to  resist  direct  compression  and  the  full  allowable  com- 
pressive strength  of  the  material  parallel  to  the  grain  may 
be  used  in  calculating  the  cross-section  required.  There- 
fore, 80,500-j-880  =  91  square  inches  is  the  sectional  area  of 
the  material  which  must  be  added  to  the  rafter  member  to 
resist  the  compressive  stress.  The  depth  of  the  rafter  being 
14  inches,  and  the  sectional  area  required  91  square  inches, 
the  width  to  add  to  the  rafter  will  be  91  -f-  14  =  6  inches. 


§  6  ARCHITECTURAL  ENGINEERING.  121 

By  combining,  the  total  width  of  the  timber  required  to 
resist  the  two  stresses  is  3^  +  04-  =  10  inches.  The  rafter 
member  will  therefore  be  made  of  one  piece  of  K)"xl4:" 
yellow-pine  timber,  which  will  extend  from  the  heel  to  the 
apex  of  the  truss. 

71.  The  tension  rods  should  be  made  of  wrought  iron 
with  eyes  formed  on  the  ends  and  provided  with  turnbuckles 
where  required ;  their  sectional  area  should  be  calculated  to 
safely  carry  the  loads  indicated  in  the  table  of  stresses. 

It  is  well  to  make  all  the  tension  bars  in  the  lower  chord 
in  pairs;  the  stresses  on  Z K,  Z J/,  and  Z Q,  Fig.  72,  will 
thus  each  be  taken  care  of  by  two  bars,  one  on  each  side  of 
the  rafter  member  and  strut. 

Since  the  stress  in  the  pair  of  members  Z  K  is  78,250 
pounds,  each  of  the  two  rods  composing  this  part  of  the 
truss  is  proportioned  so  as  to  sustain  4-  of  78,250,  or  3D,  125 
pounds.  We  will  assume  that  a  quality  of  wrought  iron  is 
used  whose  ultimate  tensile  strength  is  52,000  pounds  per 
square  inch,  and,  owing  to  the  reliability  of  the  material 
composing  them,  a  factor  of  safety  of  4  is  sufficient  in  the 
iron  tension  members;  the  allowable  tensile  strength  of  the 
wrought  iron  is,  therefore,  52, 000  -r-  4  —  13,000  pounds  per 
square  inch,  and  the  required  sectional  area  of  one  rod  is 
39, 125 -:-13, 000  =  3  square  inches.  If  square  rods  are  used, 
as  they  will  be  in  this  case,  a  If  X  If"  rod  will  be  found 
adequate,  as  it  has  a  sectional  area  of  1.75x1.75  =  3.00 
square  inches. 

The  size  of  the  other  tension  rods  may  be  found  in  a  sim- 
ilar manner,  but  as  the  stresses  upon  L  M  and  N  O,  Fig.  72, 
are  light,  these  members  may  be  made  of  a  single  tension 
bar;  these  bars  and  also  QZ  must  be  provided  with  turn- 
buckles,  which  will  be  required  to  tighten  the  whole  struc- 
ture and  take  care  of  any  slack  in  the  truss  due  to  inaccuracy 
in  construction.  In  designing  the  members  provided  with 
turnbuckles,  care  should  be  taken  to  see  that  the  rods  are  upset 
on  the  ends,  so  as  to  realize  a  sectional  area  at  the  root  of  the 
screw  thread  equal  to  the  required  sectional  area  of  the  rod. 


122 


ARCHITECTURAL  ENGINEERING. 


6 


72.  The  steel  struts  may  be  proportioned  by  the 
formula  for  calculating  the  strength  of  structural-steel  col- 
umns with  hinged  ends,  see  formula  7,  Art.  13  ;  all  the 
values  in  the  formula  are  known  except  the  square  of  the 
radius  of  gyration,  which  may  be  obtained  by  the  method 
explained  in  Art.  11.  Assume  some  convenient  section 
which  will  be  thought  to  have  the  required  resistance.  In 
this  case  the  section  shown  in  Fig.  76  is  assumed,  it  being 

convenient  for  making  the 
various  connections  to  the 
tension  rods  and  wooden 
rafter  member. 

In  order  to  calculate  the 
least  value  of  R*,  it  will  be 
-d  necessary  to  calculate  the 
least  moment  of  inertia  of 
the  section,  which  will  be 
done  in  accordance  with 
the  principles  given  in 
Art.  7. 

The  properties  of  this 
section  are  not  given  in 
the  tables,  but  the  moment  of  inertia  of  one  of  these  chan- 
nels, with  respect  to  the  axis  a  b,  is  15.47.  Since  the  neutral 
axis  a  b  of  the  section  passes  through  the  center  of  gravity 
of  the  two  channels,  the  moment  of  inertia  of  the  section, 
with  respect  to  this  axis,  is  equal  to  the  sum  of  the  moment 
of  inertia  of  the  channels,  with  respect  to  the  same  axis,  that 
is,  to  15. 47X2  =  30.94. 

The  moment  of  inertia  P  of  the  channel,  with  respect  to 
an  axis  through  its  center  of  gravity  parallel  to  c  d,  is  .  82, 
the  area  of  the  section  of  the  channel  is  3.35  square  inches, 
and  the  distance  of  its  center  of  gravity  from  the  back  of  the 
web  is  .47  inch.  The  distance  of  the  axis  through  the  center 
of  gravity  of  the  channel,  from  the  neutral  axis  c  d  of  the 
column  section,  is,  therefore,  5|-7-2  +  .47  =  3.22  inches. 
By  applying  formula  1,  Art.  7,  the  moment  of  inertia  of 
one  of  the  channels,  with  respect  to  the  axis  c  d,  is 


§  6  ARCHITECTURAL  ENGINEERING.  123 


P  =  .S2  +  3.35X3.222  =  35.55.  For  the  entire  section,  the 
moment  of  inertia,  with  respect  to  c  d,  is  equal  to  the  sum 
of  the  moments  of  its  parts,  that  is,  to  35.55x2  —  71.1. 

From  these  calculations,  it  is  evident  that  the  least 
moment  of  inertia  is  on  the  axis  ab,  and  is  equal  to  30.1)4. 
Then,  by  substituting  the  values  of  the  least  moment  of 
inertia  and  the  total  area  of  the  section  in  formula  0,  the 

square  of  the  least  radius  of  gvration  is  R*  =  •  =  4.62 

6.  70 

According  to  Table  6,  Art.  61,  Architectural  Engineering, 
§  5,  the  ultimate  compressive  strength  of  structural  steel 
is  52,000  pounds  per  square  inch;  the  length  of  the  longest 
column  in  the  truss  is  8  feet,  or  !>G  inches.  By  substituting  in 
formula  7,  Art.  13,  the  ultimate  compressive  resistance  of 

the  section  is  S  =  -  -  —  —  '—  -  -  =  46,840  pounds  per 

L  +  \18,OOOX4.62/ 
square  inch. 

Since  the  area  of  the  section  is  6.70  square  inches,  the  ulti- 
mate resistance  of  the  column  will  be  46,  840X6.  70  =  313,828 
pounds;  if  a  factor  of  safety  of  4  is  used  in  this  column,  its 
safe  resistance  will  be  313,828^-4  =  78,457  pounds. 

Since  the  stress  in  this  long  strut  or  column  is  only 
19,250  pounds,  it  can  readily  be  seen  that  it  has  several 
times  the  required  strength.  It  is,  however,  deemed  advi- 
sable to  use  this  size  of  column,  as  the  detailing  where  it 
joins  the  rafter  member  and  also  the  pin  connections  at  the 
lower  end  demand  that  channels  of  this  size  be  used. 

73.  On  account  of  the  facilities  in  making  the  connec- 
tions, and  because,  in  a  case  of  this  kind,  it  is  generally  advi- 
sable to  adopt  the  same  rolled  sections  throughout,  wherever 
possible,  the  short  struts  should   be   made  of   the  same 
rolled  sections  as  the  long  strut  ;  this  method  saves  labor  in 
the  shop  and  facilitates  assembling  and  erection  in  the  field. 

74.  The  size  of  the  pins  is  yet  to  be  determined.     This 
was  so  thoroughly  treated  in  Arts.  29-31  that  no  further 
explanation  will  be  required  here.     It  is  sufficient  to  say  that 


124  ARCHITECTURAL  ENGINEERING.  §  6 

upon  thorough  examination  of  the  several  pinned  joints,  it 
was  decided  that  a  3 -inch  pin  would  be  sufficiently  strong  to 
resist  any  bending,  shearing,  or  bearing  stresses  that  would 
be  applied  to  them. 

The  correct  design  of  the  castings  at  the  heel  and  apex  of 
the  truss  is  more  a  matter  of  experience  and  good  judgment 
than  of  calculation. 

75.  The  student  should  carefully  study  the  details  of  the 
truss,  shown  in  Fig.  77,  which  have  been  designed  according 
to  the  preceding  diagrams  and  calculations.  He  should 
observe  how  all  the  connections  are  made  and  especially  the 
details  of  the  pin  connections  and  the  castings  at  the  apex 
and  heel  of  the  truss. 

The  light  rod  at  the  center  of  the  truss  has  no  stress  on  it, 
but  is  used  simply  to  support  the  lower  central  tie-member 
and  prevent  it  from  sagging. 

In  designing  compression  members  made  up  of  two  chan- 
nels tied  together  with  plates,  as  shown  in  Fig.  77,  the  ties 
should  not  be  farther  apart  than  16  times  the  width  of  the 
flange ;  for  example,  in  this  case,  the  width  of  the  flange  of 
the  channels  composing  the  column  is  about  If  inches,  16 
times  this  width  is  28  inches ;  hence,  the  distance  between 
the  two  pieces  or  ties  should  not,  in  this  column,  be  over  28 
inches.  An  inch  one  way  or  the  other,  however,  would 
make  very  little  difference. 


THE    DESIGN    OF    A    STRUCTURAL-STEEL 
ROOF    TRUSS. 

76.  The  material  most  generally  used  in  the  construction 
of  roof  trusses  for  the  support  of  the  roof  covering  of  modern 
buildings  is  structural  steel.  The  rolled  sections  chiefly  used 
in  their  construction  are  angles  and  plates,  though  any  of 
the  other  steel  sections  may  be  adapted  to  special  cases. 

When  a  structural-steel  roof  truss  is  made  of  angles  arid 
plates,  the  angles  are  usually  connected  in  pairs  with  the 
plate  between  them,  as  shown  in  Fig.  78.  When-  this 


.u    o 
Q    J; 

s  I 


"p 

V-i. 

-*'  *. 

'T"  a 

'  —  ^^J 

'  —  • 

, 

*  1 

3  4?Ri 

,  : 

§  G  ARCHITECTURAL  ENGINEERING.  125 

construction  is  adopted,  the  joints  and  connections  of  the 
several  members  may  be  made  quite  conveniently. 

Assume  that  it  is  desirable  to  construct  the  Fink  roof  truss, 


previously  described  and  designed  as  a  pin-connected  truss, 
of  structural  steel.  The  stresses  will  be  the  same  as  given 
in  the  table,  Art.  68,  and  the  general  dimensions  as  given 
in  Fig.  72. 

The  frame  diagram  may  be  redrawn  and  the  stresses 
marked  upon  the  several  members,  as  shown  in  Fig.  70. 

77.  The  rafter  member  should  be  made  in  one  length 
from  the  heel  to  the  apex  of  the  truss  and  proportioned  to 
suit  the  greatest  stress  upon  it,  which  in  this  case  i-s  80,500 
pounds.  Generally  in  steel  construction,  the  roof  covering 
is  supported  on  purlins  which  are  placed  at  the  panel  points. 
There  is,  therefore,  no  bending  stress  in  the  rafter,  and  it  is 
subjected  to  compression  only. 

In  this  case  the  rafter  is  in  compression  only,  and  the  por- 
tion between  each  panel  point  will  be  regarded  as  a  column 
whose  length  is  equal  to  the  distance  from  center  to  center 
of  the  joints.  Assume  the  size  of  a  pair  of  angles,  which 
judgment  and  experience  dictates  as  being  adequate  to  sup- 
port the  stress  upon  the  member;  then  by  the  formula  for 


126 


ARCHITECTURAL  ENGINEERING. 


G 


structural-steel  columns  with  fixed  ends,  determine  the 
strength  of  the  assumed  section.  If  its  strength,  as  found 
by  the  formula,  is  equal  to,  or  slightly  in  excess  of,  the 
strength  required  to  resist  the  stress  in  the  member,  the  sec- 
tion may  be  adopted ;  providing,  of  course,  that  suitable  con- 
nections can  be  made  to  it. 

In  this  case  it  was  decided  to  assume  a  section  made  of 
two  5"x3^"xy  angles,  placed  back  to  Back  with  the  long 
legs  vertical,  and  about  -^  inch  apart.  The  length  of  the 
column  is  11  ft.  2^  in.,  say  134  inches;  and  from  the  table 
' '  Radii  of  Gyration  for  Two  Angles  Back  to  Back, "  the 
radius  of  gyration  of  a  section  composed  of  two  5"  X  3£"  X  f " 


-8O-O3f>an  from  C.toC- 


FlG.  79. 


angles,  with  the  long  legs  back  to  back  and  \  inch  apart,  is 
found  to  be  1.51,  a  value  sufficiently  exact  for  our  purpose. 
Substituting  these  values,  in  formula  9,  the  ultimate  com- 

pressive  strength  of  the  column  is  S  = 


\36,OOOX  1.51V 


36,OOOX  1.51 

=  42,  600  pounds  per  square  inch  of  section  ;  dividing  by  4, 
the  factor  of  safety  adopted,  the  allowable  strength  of  the 
column  per  square  inch  of  section  becomes  10,650  pounds. 
The  area  of  a  5"x3%"x¥  angle,  according  to  the  table 
"Areas  of  Angles,"  is  4  square  inches;  the  area  of  the 
entire  section  is,  therefore,  4x2  =  8  square  inches,  and  the 


§  G  ARCHITECTURAL  ENGINEERING.  12T 

allowable  strength  of  the  member,  10,650X8  =  85,200 
pounds.  Since  the  stress  on  the  member  is  only  80,500 
pounds,  it  is  evident  that  the  assumed  section  will  fulfil  the 
requirements,  and  may  be  used  for  the  rafter  member. 

78.  The  main  strut,  or  the  member  M ' N,  is  the  next 
compression  member  of  any  considerable  size;  its  length  is 
8  ft.  7  in.,  or  103  inches;  it  is  assumed  that  a  section  com- 
posed of  two  3"x2"xy  angles,  placed  back  to  back  at  a 
distance  apart  of  -^  inch,  will  suit  this  position  in  the  truss; 
the  value  of  R  for  two  3"  X  2"  X  \"  angles  placed  back  to  back 
and  l  inch  apart,  from  the  table  "  Radii  of  Gyration  for 
Two  Angles  Placed  Back  to  Back,"  is  found  to  be  .93, 
which  is  a  close  enough  approximation  for  our  purpose. 

Then  in  this  case  5  = ~ — =  39,000  pounds, 

1  + 


^30,OOOX.932; 

the  ultimate  strength  of  the  section  per  square  inch  of  area; 
since  the  area  of  the  two  3//x2'/Xi"  angles  is  2.38  square 
inches  and  the  required  factor  of  safety  is  4,  the  allowable 
strength  of  the  section  will  be  (39,000 -r- 4)  x  2. 38  =  23,200 
pounds.  Since  the  stress  in  the  member  is  only  19,250 
pounds,  it  is  evident  that  this  section  will  be  sufficiently 
strong. 

79.     The  stress  on  the  short  struts  K L  and  OP  is  so 

small  that  any  calculation  of  their  required  section  would 
only  result  in  obtaining  a  section  composed  of  such  small 
shapes  that  their  use  would  not  be  practical  in  a  truss  of 
this  size  and  character. 

Here  it  is  well  to  call  attention  to  the  practical  principle 
that  in  selecting  the  section  of  a  member  where  the  stresses 
are  light,  care  must  be  taken  to  choose  such  a  section  as  will 
best  fulfil  the  requirements  of  the  construction,  disregarding 
the  fact  that  it  will  be  stronger  than  is  actually  required  to 
sustain  the  load.  For  example,  it  is  considered  good  practice 
to  make  the  leg  of  the  angle  held  by  a  rivet  not  less  in  width 
than  three  times  the  diameter  of  the  rivet;  accordingly, 


128  ARCHITECTURAL  ENGINEERING.  §  6 

a  f -inch  rivet  should  not  be  used  in  an  angle  leg  whose  width 
is  less  than  three  times  f  inches,  or  2£  inches. 

In  the  case  of  the  truss  under  consideration,  it  was 
deemed  advisable  to  use  two  2^"  X  2"  X  \"  angles  for  each  of 
the  struts  KL  and  O  P. 

8O.  The  tension  members  in  the  truss  are  to  be  com- 
posed of  rolled  sections  similar  to  those  used  in  the  struts; 
that  is,  two  angles  back  to  back  and  far  enough  apart  to 
allow  the  T5g--inch  gusset  plate,  forming  the  means  of  con- 
nection at  the  joints,  to  slip  between  them. 

As  far  as  practicable,  tension  members,  in  common  with 
the  struts,  are  made  up  of  one  continuous  section;  thus, 
irrespective  of  the  fact  that  the  stress  in  Z  K  \&  greater  than 
in  Z  M,  these  members  should  both  be  made  of  one  contin- 
uous pair  of  angles;  by  so  doing  the  labor  is  minimized, 
thus  effecting  a  saving  that  would  more  than  offset  the  cost 
of  the  superfluous  material  in  the  member  Z  M,  besides 
producing  a  much  more  pleasing  appearance. 

In  the  truss  under  consideration,  Z K  and  ZMwill  be 
made  of  the  same  pair  of  angles;  Q  N and  QPwi\\  also  be 
composed  of  a  single  pair  of  angles;  care  being  taken  to 
proportion  the  section  to  withstand  the  greater  stress. 

The  stress  in  the  member  ZK  is  78,250  pounds;  the 
allowable  tensile  strength  of  structural  steel,  if  the  ultimate 
stress  is  taken  at  60,000  pounds,  and  a  factor  of  safety  of 
4  is  used,  will  be  60, 000  -e-  4  =  15,000  pounds  per  square 
inch.  Then  78,250-^15,000  =  5.21,  the  sectional  area  in 
square  inches  that  will  be  required  in  the  pair  of  angles 
forming  the  member  Z  K  after  deducting  the  sectional  area 
cut  out  by  one  rivet  hole  in  each  angle. 

From  the  table  "Areas  of  Angles,"  a  3"x3"X&"  angle  is 
seen  to  have  a  sectional  area  of  3.06  square  inches;  two 
angles  will  then  have  a  sectional  area  of  twice  3.06,  or  6.12 
square  inches.  Since  the  thickness  of  these  angles  is  -£$  inch 
and  a  f -inch  rivet  is  to  be  used,  the  area  of  the  section  to  be 
deducted  for  one  rivet  hole  is  .875  X  .5625  =  .49  square  inch, 
and  the  net  sectional  area  in  the  two  angles  is,  therefore, 


ARCHITECTURAL  ENGINEERING. 


129 


6. 12  —  2X.49  =  5.14  square  inches.  This  is  slightly  under 
the  sectional  area  demanded  by  the  calculation,  but,  as  a 
low  ultimate  tensile  strength  was  assumed,  it  will  be  safe 
to  use  this  section  for  the  members  Z  K  and  Z  J/. 

The  other  tension  members  in  the  truss  may  be  propor- 
tioned in  the  same  manner,  care  being  taken  to  deduct  from 
the  section  area  of  the  member  the  sectional  area  removed 
for  rivet  holes;  also,  that  the  rolled  sections  adopted  will 
satisfy  the  practical  demands  of  the  construction. 

81.  The  number  of  rivets  required  at  the  several  joints 
in  the  truss  should  be  carefully  calculated.  Since  the 
method  of  calculation  is  the  same  for  all  the  joints,  only 
one  will  be  considered  here,  though  the  student  should 
analyze  the  others  for  himself. 


FIG.  80. 

Fig.  80  represents  a  detail  of  the  joint  at  a,  Fig.  70.  The 
stress  in  the  member  M ' N 'is  19,250  pounds,  and  a  sufficient 
rivet  section  must  be  provided  at  the  end  of  this  member 
to  resist  this  stress.  By  referring  to  the  table  "Values  of 
Rivets,"  and  using  an  allowable  stress  per  square  inch  of 
15,000  pounds,  the  least  value  of  a  f-inch  rivet  in  this  con- 
nection is  found  to  be  the  web  bearing  of  a  fV-inch  Plate» 
which  is  6,094  pounds;  then  19, 250 -=-6, 094  ==  3.15,  or  say 
4,  the  number  of  rivets  required  in  this  connection. 
1  so 


130  ARCHITECTURAL  ENGINEERING.  §  6 

The  member  N  Q  is  subjected  to  a  stress  of  34, 000  pounds, 
the  least  value  for  the  rivets  is  the  same  as  before,  and  the 
number  required  in  the  end  of  this  member  is  34,000-^6,094 
=  5. 56,  say  6  rivets. 

The  stress  in  the  member  ZM  is  64,000  pounds,  which 
will  require  a  large  number  of  rivets.  Using  8  rivets  in  the 
vertical  legs  and  gusset  plate,  as  shown  in  Fig.  80,  the  value 
of  each  is  6,094  pounds  and  the  value  of  eight  is  6,094x8 
=  48,752  pounds.  This  deducted  from  64,000  pounds,  the 
entire  stress  in  the  member,  leaves  15,248  pounds  yet  to  be 
provided  for  by  the  use  of  a  splice  plate  attached  to  the 
horizontal  legs  of  the  angles.  The  value  of  one  rivet  in  the 
horizontal  legs  of  the  angles  forming  the  member  Q  Z  is  the 
ordinary  bearing  value  of  a  f -inch  rivet  in  a  ^-inch  plate, 
which,  according  to  the  table  "Values  of  Rivets,"  is  3,656 
pounds,  with  an  allowable  stress  of  15,000  pounds.  The 
value  of  four  rivets  will  then  be  3,656x4  =  14,624  pounds, 
a  little  less  than  the  amount  to  be  taken  care  of,  but  the 
difference  is  so  small  that  it  may  be  disregarded  and  for  the 
sake  of  symmetry  the  same  number  of  rivets  will  be  used  in 
connecting  the  .splice  plate  to  the  member  M  Z.  The  num- 
ber of  rivets  required  to  connect  the  member  Z  Q  to  the 
gusset  plate  may  be  readily  obtained. 

82.  Fig.  81  shows  the  usual  shop  drawing  of  the  truss 
just  designed;  the  student  should  pay  particular  attention 
to  the  details  of  the  connections.  Separators  should  be 
placed  in  both  the  tension  and  compression  members;  they 
are  placed  in  the  tension  members  to  prevent  the  angles 
from  striking  against  each  other  when  the  trusses  are  sub- 
jected to  vibrations,  and  also  to  join  the  two  angles  of  a 
member,  so  that  the  work  will  arrive  at  the  point  of  erection 
in  a  convenient  form,  ready  to  put  together.  The  separators 
are  placed  in  the  compression  members  so  as  to  insure 
against  any  tendency  to  bend  them  apart,  and  so  that  they 
will  act  in  unison.  The  spacing  of  these  separators  is  more 
a  matter  of  judgment  on  the  part  of  the  designer  than  any- 
thing else,  though  any  spacing  over  8  times  the  least 


\5eparators £  th/ck. 


A//  Rivets  $  diam.  unless 
otfier-wise  marked. 
AllGussef  Plates 


P/9ce  Separators  where  shown. 


SO/I 'from  Center  to  Center  of  Jnc/ror  Softs.  - 


§  6  ARCHITECTURAL  ENGINEERING.  131 

dimension  of  the  member  is  not  to  be  recommended.  Separa- 
tors should  always  be  placed  near  the  end  of  a  pair  of  angles  to 
be  connected  to  a  gusset  plate,  as  it  will  hold  them  the  right 
distance  apart  in  shipment  and  facilitate  erection  in  the  field. 
The  lf"Xlf"X  V  angle,  joining  the  apex  of  the  truss  and 
its  lower  chord,  is  required  only  to  support  the  chord  angles 
and  prevent  them  from  sagging.  When  there  is  consider- 
able stress  in  this  member,  sagging  is  unlikely  to  occur,  but 
it  is  the  usual  practice,  and  a  good  one,  to  introduce  some 
support  for  this  long  and  usually  light  member.  It  will  be 
noticed  that  one  size  of  rivets  is  used  throughout  this  truss 
wherever  practicable ;  this  is  a  point  in  economical  construc- 
tion that  should  always  be  observed.  The  ends  of  the  angles 
and  members,  when  practicable,  should  be  cut  off  square, 
and  the  gusset  plates  should  be  designed  so  that  they  may 
be  formed  with  as  few  cuts  as  possible,  and  unless  it  is  desired 
to  make  the  truss  somewhat  fanciful,  these  cuts  should  never 
be  other  than  straight  lines. 

In  designing  roof  trusses,  and,  in  fact,  all  structural-steel 
constructions,  care  should  be  taken  to  see  that  the  several 
sections  into  which  each  may  be  divided  are  not  so  large  that 
they  cannot  be  shipped  by  railroad  or  other  transportation 
at  hand ;  this  is  important  and  should  be  carefully  considered. 


GENERAL  NOTES  REGARDING  THE  DESIGN  OF 
A  ROOF  TRUSS. 

83.  Lateral  Bracing. — Trusses  forming  the  principal 
support  of  a  roof,  if  of  any  considerable  size,  should  be 
braced  together  in  the  planes  of  the  rafters  so  as  to  secure 
them  against  any  tendency  of  the  wind,  when  blowing  in  a 
direction  perpendicular  to  the  gable  ends,  to  produce  lateral 
movement.  If  the  roof  sheathing  is  laid  close  and  is  well 
nailed,  it  will  sufficiently  stiffen  trusses  of  moderate  span. 
The  heels  of  trusses  are  sometimes  fastened  securely  to  the 
walls,  especially  in  those  buildings  where  the  wind  is  liable 
to  get  under  the  roof.  When  so  secured  there  is  a  tendency 


132  ARCHITECTURAL  ENGINEERING.  §  6 

for  the  wind  to  reverse  the  stresses  in  the  members  of  a  roof 
provided  with  a  light  covering,  and  this  reversal  should  be 
provided  for  in  the  design  of  the  truss. 

84.  Factors  of  Safety. — Since  due  allowance  must  be 
made  for  unforeseen  and  unknown  defects  of  material  and 
workmanship,  and  for  unknown  stresses  which  are  liable  to 
occur,  it  is  necessary  to  proportion  the  several  parts  of  a 
structure  so  that  they  will  be  able  to  resist,  without  failure, 
much  larger  forces   than   those   obtained   from  the  stress 
diagram. 

In  roof  trusses,  however,  the  stresses  can  be  calculated 
with  more  certainty  than  in  the  case  of  a  bridge  or  a  machine, 
and  their  application  is  more  steady  in  its  nature,  and,  there- 
fore, not  so  severe  on  the  material.  For  this  reason  it  is 
permissible  to  allow  unit  stresses,  in  the  design  of  roof 
trusses,  some  fifty  per  cent,  in  excess  of  those  considered 
allowable  in  first-class  bridges. 

85.  Tension    Members. — The    strength    of   a   tension 
member  is  that  of  the  smallest  cross-section.     It  is,  therefore, 
good  practice  to  upset  or  enlarge  the  ends  of  long  bolts  or 
tension  rods  with  screw  ends,  so  that  the  cross-section  at  the 
root  of  the  thread  will  be  at  least  equal  to  the  sectional  area 
of  the  main  part  of  the  rod. 

The  central  axis  of  the  cross-section  of  ties  and  struts 
should  coincide  with  the  line  of  action  of  the  thrust  or  pull, 
as  otherwise  the  piece  will  be  greatly  weakened  and  danger- 
ous bending  moments  will  be  developed  in  the  structure. 
To  calculate  the  net  sectional  area  required  in  any  tension 
member,  the  force  or  load  upon  it  should  be  divided  by  the 
safe  working  tensile  strength  of  the  material,  and  allowance 
must  be  made  in  the  area  of  the  cross-section  for  the  cutting 
of  bolt  and  rivet  holes. 

86.  Compression    members   whose    lengths    do    not 
exceed  six  times  the  least  dimension  may  be  proportioned 
by   the    method  followed   for  the   tie   as  just    explained; 


§  6  ARCHITECTURAL  ENGINEERING.  133 

but,  when  the  length  of  the  truss  is  increased,  there  is  a 
tendency  to  yield  sidewise  when  compressed,  and  the  sec- 
tional area  must  be  increased  or  the  unit  stress  diminished. 
Hence,  the  column  formulas  given  in  Arts.  13-15  must  be 
used.  Pieces  subjected  to  alternate  compression  and  ten- 
sion should  have  a  materially  larger  section  than  would  be 
required  for  either  stress  alone.  Cast  iron  is  seldom  used 
in  the  best  work  for  anything  but  short  compression  pieces, 
packing  blocks,  and  pedestals. 

87.  Members  in  Trusses   Subjected  to   Transverse 
Stresses. — In  determining  the  resisting  moment  of  a  mem- 
ber subjected  to  transverse  stresses,  or  in   calculating  the 
section  required  at  the  point  of  maximum  bending  moment, 
due  allowance  must  be  made  for  portions  cut  away  on  the 
tension  side  in  attaching  fastenings,  or  in  making  connec- 
tions; similar  allowance  must  also  be  made  on  the  compres- 
sion side,  unless  the  holes  are  completely  filled  by  the  rivets, 
in  which  case  no  deduction  need  be  made  from  the  sectional 
area  of  the  member. 

88.  Members  in  Trusses  Subjected  to  IJotli  Trans- 
verse and  Direct  Stresses. — The  rafter  members  in  a  roof 
truss,  likewise  members  in  other  structures,  are  often  called 
upon  to  resist  both  a  bending  and  a  direct  stress.     Such 
pieces  must  first  be  designed  to  safely  resist  the  bending 
moment,  and    then   their   transverse   dimensions   must   be 
increased  so  that  the  added  material  will  have  sectional  area 
sufficient  to  resist  the  direct  pull  or  thrust.     Should   the 
direct  force  be  a  compressive  stress,  it  will  be  well  to  test 
the  size  of  the  piece  by  the  proper  column  formula. 

89.  Pins  and  Eyes. — In  proportioning  the  pins  and  eyes 
of  tension  bars,  the  diameter  of  the  pin  should  be  from  three- 
fourths  to  four-fifths  of  the  width  of  the  bar  in  flats,  and  one 
and  one-fourth   times  the  diameter  of  the  bar  in   rounds. 
The  sectional  area  of  the  metal  around  the  eye  should  be 
fifty  per  cent  in  excess  of  that  of  the  rod  or  bar.     When 


134  ARCHITECTURAL  ENGINEERING.  §  6 

flat  bars  are  used,  their  thickness  should  be  not  less  than 
one-fourth  of  their  width ;  this  will  secure  a  good  bearing 
surface  on  the  pin.  The  size  of  a  pin  is  usually  decided  by 
the  bending  moment  on  it,  consequently  the  assembled 
pieces  on  a  pin  should  be  packed  close  together,  and  oppo- 
sing members  should  be  brought  as  nearly  in  line  with  one 
another  as  possible. 

9O.  Details. — The  designer  should  carefully  examine 
each  joint  and  connection  in  the  structure,  and  consider  such 
practical  points  as  means  of  shipment,  erection  in  the  field, 
etc.  Care  should  be  taken  in  designing  all  connections,  to 
so  place  the  rivets  and  bolts  as  to  realize  their  full  strength, 
and  at  the  same  time  not  cut  away  too  much  of  the  material 
of  the  members  connected.  All  members  and  joints  should 
be  examined  for  tension,  compression,  shearing,  and  bending, 
and  proportioned  accordingly.  The  strength  of  the  joints 
and  connections  between  the  members  of  a  structure  are  of 
as  much  importance  as  the  strength  of  the  members  them- 
selves; the  strength  of  any  structure  depends  upon  the 
strength  of  its  weakest  point,  and  its  failure  at  a  joint  or 
connection  is  as  fatal  as  the  failure  of  any  of  its  members. 

The  student  will  find  the  handbooks  issued  by  the  various 
steel  mills  of  great  value  to  him  in  the  prosecution  of  his 
work.  They  contain  many  useful  tables  giving  the  proper- 
ties of  rolled  sections,  with  information  as  to  their  use  and 
application. 


ELEMENTS    OF   USUAL    SECTIONS. 

T,Momentye3fe-r  t0  horizontal  axis  through  center  of  eravitv 
This  table  is  intended  tor  convenient  application  where  extreme  accJl 

racy  is  not  important.  The  values  for  the  last  seven  sections  and  those 
marked  »  are  approximate.  A  =  area  of  section;  in  case  of  hollow 
section,  a  =  area  of  interior  space. 


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Moment  of 
Inertia. 
/ 

Section 
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from  Cente 
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ARCHITECTURAL  ENGINEERING. 


137 


ATCEAS    OF    AXGI/ES. 

WITH  EQUAL  LEGS. 


Size  in 

Thickness  in  Inches. 

Inches. 

I 

T5* 

3                 7 
S              Tff 

* 

g 

1 

11 

3 

1  8 
Iff 

1        1 

6X6 

4.365.00 

5.75 

0.437.11 

7.78!8.449.0o'9.74  11.0 

5X5 

3  .  61  4  .  18  4  .  75  5  .  31  5  .  86  0  .  420  .  94  7  .  47  7  .  99    9  .  0 

4X4 

2.863.31 

3.754.184.615.035.445.84 

Ql   v  31 
O-j  A  O  9- 

2.48!2.87 

3.253.023.994.344.095.03 

3X3" 

1.44 

1  .  78  2  .112.  43  2  .  75  3  .  00  3  .  303  .  05 

2fx2f 

1.31 

1.62 

1.922.21 

2.50 

91  v  9  1 

*v"5~  /\  &~n 

1.19 

1.47  1.732.002.25 

2^X2^ 

1.06 

1.31 

1.55  1.782.00 

2X2 

0.94 

1.15 

1.36 

1.50 

ifxif 

0.81 

1.00 

1.17 

1  .  30 

1  i  v  1  i 
A¥  A  1^f 

0.69 

0.84 

0.99 

WITH  UNEQUAL  LEGS. 

7X31 

4.405.00 

5.59 

6.176.757.317.87 

8  .  42  9  .  50 

6X4 

3.61 

4.184.75 

5.30 

5.866.41 

6.947.477.999.00 

6X3£ 

3.42 

3.96 

4.50 

5.03 

5.556.06 

6.507.007.55S.;>0 

5X4 

3.23 

3  .  74  4  .  25 

4.74 

5.235.72 

6.186.65 

7.118.00 

5X3^ 

3.05 

3.52 

4.00 

4.464.925.375.816.25 

6.67 

5X3 

2.86 

3.303.75 

4.174.605.035.445.84 

4|X3 

2.67 

3.093.50 

3.904.304.685.005.43 

4X3^ 

2.67 

3.093.50 

3.904.304.685.065.43 

4X3 

2.09 

2.48 

2.873.25 

3.623.984.344.095.03 

3|X3 

1.93 

2.30 

2.653.00 

3.343.67 

4.00 

4.31 

4.62 

3^X2^ 

1.44 

1.782.11 

2.432.75 

3.063.36 

3.65 

3X2| 

1.31 

1.62 

1.92 

2.21 

2.50 

2.78 

3X2 

1.19 

1.46 

1.73 

1.99 

2.25 

2^X2 

1.06 

1.31 

1.55 

1.78 

2.00 

2|-Xl|- 

0.88 

1,07 

1.27 

1.45 

1.63 

2Xlf 

0.78 

138 


ARCHITECTURAL  ENGINEERING. 


PROPERTIES    OF    ANGLES. 

EQUAL  LEGS. 
c 


0) 

. 

c 

a 

1/5 

| 

d 

.2 

c  S  S> 

0)   O   W> 

1 

o 
1   • 

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fa 

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c  ^ 

^  p\ 

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>rS 

"5: 

£3 

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o 

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t[  t  ^* 

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CD  ^) 

a 

a. 

CC 

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g 

o 

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VM 

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0)  as  ° 

n  'S 

Uj'S 

tn'« 

5 

h 

'S 

01 

0 

d0  cj 

|< 

^< 

T3 

^ 

«jj 

"5:  "o  P3 

o 

at 

3 

5 

" 

K 

M 

In. 

In. 

Lb. 

Sq.  In. 

In. 

^ 

^ 

R' 

6X6 

| 

33.1 

9.74 

1.82 

31.92 

1.81 

1.17 

6X6 

17.2 

5.06 

1.66 

17.68 

1.87 

1.19 

5X5 

7 
"8 

27.2 

7.99 

1.57 

17.75 

L49 

0.98 

5X5 

3 

¥ 

•  12.3 

3.61 

1.39 

8.74 

1.56 

0.99 

4x4 

1  3 
1  6 

19.9 

5.84 

1.29 

8.14 

1.18 

0.80 

4x4 

fV 

8.2 

2.40 

1.12 

3.71 

1.24 

0.82 

3|x3i 

1  3 
T6" 

17.1 

5.03 

1.17 

*5.25 

1.02 

0.69 

31   V  31 
O-jj-  A  Ojr 

3 

8.5 

2.48 

1.01 

2.87 

1.07 

0.70 

3X3 

f 

11.4 

3.36 

0.98 

2.62 

0.88 

0.59 

3X3 

4.9 

1.44 

0.84 

1.24 

0.93 

0.60 

2fX2f 

1. 

8.5 

2.50 

0.87 

1.67 

0.82 

0.54 

2fx2f 

1 

4.5 

1.31 

0.78 

0.93 

0.85 

0.55 

2£x2| 

* 

7.7 

2.25 

0.81 

1.23 

0.74 

0.49 

01  v  9  1 

IT  ^     "2" 

i 

4.1 

1.19 

0.72 

0.70 

0.77 

0.50 

2^X2^ 

f 

6.8 

2.00 

0.74 

0.87 

0.66 

0.48 

2^X2£ 

i 

3.7 

1.06 

0.66 

0.51 

0.69 

0.46 

2X2 

rV 

5.3 

1.56 

0.66 

0.54 

0.59 

0.39 

2X2 

_3 

2.5 

0.72 

0.57 

0.28 

0.62 

0.40 

ARCHITECTURAL  ENGINEERING. 


139 


PROPERTIES   OF  ANGLES—  Continued. 
EQUAL  LEGS. 


oi 

"o 
o 

o 

.2 

S-i 

-H  c   • 
gg| 

.2" 

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fll      > 

s" 

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fa 

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CC 

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o  ^ 

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a 

o 

2 

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CD    rt^O 

a"S 

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w  * 

05  J*j 

5 

£H 

'5 

aj 

CD 

rt°  rt 

S^ 

^ 

S 

£ 

< 

.w'gW 

o 

aj 

ClJ 

p 

^ 

PH 

* 

In. 

In. 

Lb. 

Sq.  In. 

In. 

^ 

* 

A" 

ifxif 

_!_ 

4.6 

1.30 

0.59 

0.350 

0.51 

0.35 

ifxif 

_3 

2.1 

0.62 

0.51 

0.180 

0.54 

0.36 

Hxi£ 

I 

3.4 

0.99 

0.51 

0.190 

0.44 

0.31 

l|Xl| 

3 

T6 

1.8 

0.53 

0.44 

0.110 

0.46 

0.32 

lixii 

A 

2.4 

0.69 

0.42 

0.090 

0.36 

0.25 

i^xii 

i 

1.0 

0.30 

0.35 

0.044 

0.38 

0.26 

1X1 

i 

1.5 

0.44 

0.34 

0.037 

0.29 

0.20 

1X1 

i 

0.8 

0.24 

0.30 

0.022 

0.31 

0.21 

fxf 

A 

1.0 

0.29 

0.29 

0.019 

0.26 

0.18 

i 

0.7 

0.21 

0.26 

0.014 

0.26 

0.19 

fxf 

A 

0.8 

0.25 

0.26 

0.012 

0.22 

0.16 

fxf 

i 

0.6 

0.17 

0.23 

0.009 

0.23 

0.17 

140 


ARCHITECTURAL  ENGINEERING. 


PROPERTIES    OF    ANGLES. 

UNEQUAL  LEGS. 


Dimen- 
sions. 

Thickness. 

IH 

Oi    . 

-"-•  "o 
'3 

Area  of  Sec- 
tion. 

Distances 
of  Center 
of  Gravity 
from  Back 
of  Flange. 

Moments  of 
Inertia. 
/. 

Radii  of  Gyra- 
tion.    R. 

Axis 
AB. 

Axis 
CD. 

oj  *fH  ctf 

Axis 
AB. 

Axis 
CD. 

In. 

In. 

Lb. 

Sq.In. 

</. 

d* 

6X4 

1 

27.2 

7.99 

2.12 

1.12 

27.73 

9.75 

1.86 

1.11 

.88 

6X4 

I 

12.3 

3.61 

1.94 

0.94 

13.47 

4.90 

1.93 

1.17 

.88 

6X31 

1 

25.7 

7.5'5 

2.22 

0.97 

26.38 

6.55 

1.87 

0.93 

.78 

6X31 

I 

11.7 

3.42 

2.04 

0.79 

12.86 

3.34 

1.94 

0.99 

.77 

5X4 

7 
¥ 

24.2 

7.11 

1.71 

1.21 

16.42 

9.23 

1.52 

1.14 

.88 

5X4 

I 

11.0 

3.23 

1.53 

1.03 

8.14 

4.67 

1.59 

1.20 

.86 

5X31 

| 

22.7 

6.67 

1.79 

1.04 

15.67 

6.21 

1.53 

0.96 

.77 

5X31 

3 

8" 

10.4 

3.05 

1.61 

0.86 

7.78 

3.18 

1.60 

1.02 

.76 

5X3 

13 

IT 

19^9 

5.84 

1.86 

0.86 

13.98 

3.71 

1.55 

0.80 

.66 

5X3 

TT 

8.2 

2.40 

1.68 

0.68 

6.26 

1.75 

1.61 

0.85 

.66 

4^X3 

13 

T6" 

18.5 

5.43 

1.65 

0^90 

10.33 

3.60 

1.38 

0.81 

.67 

41X3 

f 

9.1 

2.67 

1.49 

0.74 

5.50 

1.98 

1.44 

0.86 

.66 

4x3^ 

« 

18.5 

5.43 

1.36 

1.11 

7.77 

5.49 

1.19 

1.01 

.74 

4x31 

t 

9.1 

2.67 

1.21 

0.96 

4.18 

2.99 

1.25 

1.06 

.73 

4X3 

yf 

17.1 

5.03 

1.44 

0.94 

7.34 

3.47 

1.21 

0.83 

.66 

4X3 

7.1 

2.09 

1.26 

0.76 

3.38 

1.65 

1.27 

0.89 

.65 

31x3 

H 

15.7 

4.62 

1.23 

0.98 

4.98 

3.33 

1.04 

0.85 

.65 

31X3 

fV 

6.6 

1.93 

1.06 

0.81 

2.33 

1.58 

1.10 

0.90 

.63 

31X21 

V 

12.4 

3.65 

1.27 

0.77 

4.13 

1.72 

1.06 

0.67 

.58 

31X21 

*  T 

4.9 

1.44 

1.11 

0.61 

1.80 

0.78 

1.12 

0.74 

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3^X2 

_^. 

9.0 

2.64 

1.21 

0.59 

2.64 

0.75 

1.00 

0.53 

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3^X2 

i 

4.3 

1.25 

1.09 

0.48 

1.36 

0.40 

1.04 

0.57 

.44 

o  \/  o  i 

*J  /\  &~*F 

IT 

9.5 

2.78 

1.02 

0.77 

2.28 

1.42 

0.91 

0.72 

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3X21 

i 

4.5 

1.31 

0.91 

0.66 

1.17 

0.74 

0.95 

0.75 

.53 

ARCHITECTURAL  ENGINEERING. 


HI 


PROPERTIES    OF   ANGI/KS—  Continued. 
UNEQUAL  LEGS. 


t/i 

in 

CD 

o 

CD 

02 

Distances 

Moments  of 

Radii  of  Gyra- 
tion.    A'. 

Dimen- 

a 

58 

•4-1     ,4 

0    0 

of  Gravity 

Inertia. 
/ 

'S   >r  T-4 

sions. 

o 

bC£n 

from  Back 

oj  'p   « 

^ 

CD 

CD  *"" 

of  Flange. 

Axis 

Axis 

*<J    0 
•i-1        bo 

•^ 

"^ 

Axis 

Axis 

A  B. 

C  D. 

i  ••  rt 

In. 

In. 

Lb. 

Sq.In. 

d, 

dt 

A  />'. 

CD. 

J  ^Q 

3X2 

j. 

7.7 

2.25 

1.08 

0.58 

1.92 

0.07 

0.92 

0.55 

.47 

3X2 

¥ 

4.0 

1.19 

0.99 

0.49 

1.09 

0.39 

0.95 

0.50 

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2|X2 

1 

•2 

6.8 

2.00 

0.88 

0.03 

1.14 

0.04 

0.75 

0.50 

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2^X2 

3 
1  « 

2.8 

0.81 

0.70 

0.51 

0.51 

0.29 

0.79 

0.00 

.43 

2^X11 

1 

2 

5.5 

1  .  03 

0.80 

0.48 

0.82 

0.20 

0.71 

0.40 

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2iXl| 

3 
1  (i 

2.3 

0.07 

0.75 

0.37 

0.34 

0.12 

0.72 

0.43 

.40 

2X1| 

1 
4 

2.7 

0.78 

0.69 

0.37 

0.37 

0.12 

0.03 

0.39 

.  30 

2X1| 

JL 

2.1 

0.00 

0.66 

0.35 

0.24 

0.09 

0.0:5 

0.40 

.29 

PROPERTIES    OF    Z    BARS. 


JS 


I 


I 


oi    . 

• 

VM     £  _. 

CD 

O   CD 

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tn 

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CD 

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rt  S'S 

Q 

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In. 

In. 

Lb. 

Sq.In. 

' 

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15.6 

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25.32 

2.35 

9.11 

1.41 

0.83 

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5.  39 

29.80 

2.35 

10.95 

1.43 

0.84 

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34.36 

2.36 

12.87 

1.44 

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6.68 

34.  64 

2.28 

12.59 

1.37 

0.81 

6yV 

3y97 

5 
T 

25.4 

7.46 

38.86 

2.28 

14.42 

1.39 

0.  82 

H 

3f 

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28.0 

8.25 

43.18 

2.29 

16.34 

1.41 

0.84 

142 


ARCHITECTURAL  ENGINEERING. 


PROPERTIES    OF  Z  BARS — Continued. 


ca 

p 

*tt   5?f     ' 

C    -j  '  —  ' 

o 

*o 

;-. 

<*-! 

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M-i 

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I 

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f 

29.3 

8.63 

42.12 

2.21 

15.44 

1.34 

0.81 

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1  3 

32.0 

9.40 

46.13 

2.22 

17.27 

1.36 

0.82 

61 

3| 

7 

¥ 

34.6 

10.17 

50.22 

2.22 

19.18 

1.37 

0.83 

5 

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11.6 

3.40 

13.36 

1.98 

6.18 

1.35 

0.75 

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4.10 

16.18 

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16.4 

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19.07 

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q  5 

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9 

20.2 

5.94 

21.83 

1.91 

10.51 

1.33 

0.75 

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¥ 

22.6 

6.64 

24.53 

1.92 

12.06 

1.35 

0.76 

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31 

1  1 

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23.7 

6.96 

23.68 

1.84 

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11.4 

3.36 

4.57 

1.17 

4.75 

1.19 

0.56 

3 

2j| 

% 

12.5 

3.69 

4.59 

1.12 

4.85 

1.15 

0.55 

3yV 

2f 

A 

14.2 

4.19 

5.26 

1.12 

5.70 

1.17 

0.56 

ARCHITECTURAL   ENGINEERING. 


PROPERTIES    OF    I    BEAMS. 

-f— I- 


143 


-      x 

f  <          £< 


In. 

Lb. 

Sq.  In. 

In. 

In. 

/ 

/ 

K 

20 

90 

20.4 

.78 

0.7-T 

1.5iM.5  .  lo    7  .55 

42.30 

1  .27 

20 

80 

23  .  5 

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i;  .  38 

1.345.10    7.55 

33  .  2<  ' 

1.19 

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1.240.90    7.53 

28.20 

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6.29 

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178.50    3.89 

10.30 
13.50 

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1.07 

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9.7 

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5  .  00 

101.30    4.08 

11.80 

1.10 

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4.89 

134.50    3.90 

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0  .  96 

10 

25 

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4.75 

122.50   4.00 

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7.9 

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4.75 

110.60    3.72 

9.10 

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5.56 

0  .  95 

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27 

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6.91 

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69.70    3.30 

6.02 

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> 

18 

5  .  2 

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56.80   3.30 

3.95 

0.87 

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47.60    2.89 

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0  .  92 

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5.86    1.63 
4.59    1.61 

0.70 
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144 


ARCHITECTURAL  ENGINEERING. 


PROPERTIES    OF    CHANNELS. 


J 


B 


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c    • 

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371.60 

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1.76 

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6.53 

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5 

1.46 

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1.59 

3.59 

1.57 

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ARCHITECTURAL  ENGINEERING. 


RADII    OF    GTRATION    FOR    TWO   ANGLES. 

PLACED  BACK  TO  BACK,  SHORT  LEG  VERTICAL. 

_ *j ^ 5*__i_  _«. 

^  ^•MMB  MHMMT  W^M   BP^MMBr  *— _ 

!      |>          *«      W          *!     * 

Iff 

UNEQUAL  LEGS. 

77^r  different  radii  of  gyration  arc  indicated  in  the  figures 
by  arrowheads. 


Size. 
Inches. 

Thickness. 
Inches. 

Radii  of  Gyration. 

R> 

A\ 

X, 

X. 

6X4 

1 

1.19 

2.94 

3.13 

3.23 

6X4 

3 

1.17 

2.74 

2.92 

3.02 

5X31 

1 

1.01 

2.39 

2.58 

2.68 

5x31 

3 

¥ 

1.02 

2.27 

2.45 

2.55 

5X3 

f 

.86 

2.50 

2.69 

2.79 

5X3 

A 

.85 

2.33 

2.51 

2.61 

41X3 

i 

.86 

2.18 

2.38 

2.46 

41X3 

5 
TT 

.87 

2.06 

2.25 

2.33 

4X31 

1 

1.05 

1.85 

2.04 

2.14 

4X31 

5 
TT 

1.07 

1.73 

1.91 

2.00 

4X3 

5 

.83 

1.84 

2.03 

2.13 

4x3 

A 

.89 

1.79 

1.97 

2.07 

31X3 

5 

.87 

1.57 

1.76 

1.87 

31X3 

A 

.90 

1.53 

1.71 

1.81 

31X21 

iV 

.72 

1.66 

1.85 

1.95 

31X21 

i 

.74 

1.58 

1.76 

1.86 

3X21 

TV 

.73 

1.40 

1.59 

1.69 

3X21 

i 

.75 

1.32 

1.49 

1.60 

3X2 

.55 

1.42 

1.62 

1.72 

3X2 

1 

.57 

1.39 

1.57 

1.68 

1-31 


146  ARCHITECTURAL  ENGINEERING.  §6 

RADII    OF    GYRATION    FOR    TWO   ANGLES. 

PLACED  BACK  TO  BACK,  LONG  LEG  VERTICAL. 

R .  R*  It* 


T>*    r~   i"~i  f~~ 
1  tt     1  TT 

-I|T-  3fK- 

UNEQUAL  LEGS. 

The  different  radii  of  gyration  are  indicated  in  the  figures 
by  arrowheads. 


Size. 
Inches. 

Thickness. 
Inches. 

Radii  of  Gyration. 

R» 

A 

R* 

R3 

6X4 

7 

1.95 

1.68 

1.87 

1.97 

6X4 

3 

1.93 

1.50 

1.67 

1.76 

5X31 

t 

1.59 

1.44 

1.63 

1.73 

5  X  3i- 

3 

1.60 

1.34 

1.51 

1.61 

5X3 

3 
¥ 

1.62 

1.23 

1.42 

1.52 

5X3 

* 

1.61 

1.09 

1.26 

1.36 

4|x3 

3 

T 

1.43 

1.25 

1.44 

1.55 

4|X3 

TT 

1.45 

1.13 

1.31 

1.40 

4X31- 

3 

¥ 

1.24 

1.53 

1.72 

1.83 

4X3£ 

5 

1.26 

1.41 

1.58 

1.69 

4X3 

Y 

1.23 

1.20 

1.39 

1.50 

4X3 

5 

TB" 

1.27 

1.17 

1.35 

1.45 

31x3 

I 

1.06 

1.27 

1.46 

1.56 

3^-X3 

5 
Tfr 

1.10 

1.21 

1.39 

1.49 

Q  1  v  9  1 
t»2"  S\  <VY 

1.10 

1.04 

1.23 

1.34 

Q  1  \s  O  1 
O-j  s\  *~2" 

¥ 

1.12 

.96 

1.17 

1.24 

3x2£ 

T^ 

.93 

1.07 

1.27 

1.37 

3x2£ 

i 

.95 

1.00 

1.18 

1.28 

3X2 

.92 

.80 

1.00 

1.10 

3X2 

! 

.96 

.75 

.93 

1.04 

HXH 

f5f 

.70 

.60 

.79 

.91 

2iXli 

A 

.72 

.57 

.75 

.86 

ARCHITECTURAL  ENGINEERING. 


14? 


RADII    OF    GYRATIOX    FOR    TWO    AXGLES. 

PLACED  BACK  TO  BACK. 


EQUAL  LEGS. 

The  different  radii  of  gyration  arc  indicated  in  t lie  figures 
by  arrowheads. 


Size. 
Inches. 

Thickness. 
Inches. 

Radii  of  Gyration. 

A\, 

*, 

*, 

ft* 

6X6 

| 

1.87 

2  .  64 

2.83 

2.92 

6X6 

3 

¥ 

1.88 

2.49 

2.66 

2.75 

5X5 
5X5 

8 

¥ 
8 

¥ 

1.55 
1.56 

2.20 
2.09 

2  .  38 

2.27 

2.48 
2.36 

4X4 
4x4 

1  3 
T5" 

1.24 
1.24 

1  .  83 
1.67 

2.03 
1.85 

2.12 
1  .  94 

31x1 

5 

¥ 

1.04 
1.08 

1.51 
1.46 

1  .  70 
1.65 

1.81 
1.74 

3X3 
3X3 

5 

.94 
.93 

1.40 
1.25 

1.59 
1.43 

1.69 
1.53 

*ixSt 

1 
* 

.76 

.77 

1.12 
1.05 

1.31 
1.25 

1.42 
1  .  34 

21  X21 

1 

.70 

1.05 

1.25 

1.35 

2jx2i 

A 

.69 

.94 

1.12 

1.22 

2X2 

2X2 

A 

.62 
.62 

.95 

.84 

1.15 
1.03 

1.26 
1.13 

148  ARCHITECTURAL  ENGINEERING. 


MODULI    OF    ELASTICITY. 


METALS. 


Iron  (cast), 12,000,000 

Iron  (wrought  shapes),  ......  27,000,000 

Iron  (rerolled  bars),  .     .     .     .     .     .     .  26,000,000 

Steel  (casting), 30,000,000 

Steel  (structural), 29,000,000 

TIMBER. 

Chestnut, 1,000,000 

Cypress, 900,000 

Cedar, 700,000 

Hemlock, 900,000 

Oak  (White), 1,100,000 

Pine  (White), 1,000,000 

Pine  (Southern,  Long-leaf,  or  Georgia 

Yellow  Pine), 1,700,000 

Pine  (Douglass,  Oregon  and  Washing- 
ton Fir,  or  Yellow  Pine),       .     .     „  1,400,000 
Pine     (Northern     Short-leaf     Yellow 

Pine), 1,200,000 

Pine  (Red),       . 1,200,000 

Pine  (Norway), 1,200,000 

Pine  (Red,  Ontario,  Canadian),  .     .     .  1,400,000 

Redwood  (California), 700,000 

Spruce  and  Eastern  Fir, 1,200,000 

Spruce  (California), 1,200,000 


ARCHITECTURAL  ENGINEERING. 


RESISTING    MOMENTS    OF    PINS. 

WITH  EXTREME  FIBER  STRESSES 
VARYING  FROM  15,000  TO  25,000  POUNDS  PER  SQUARE  INCH. 


Diame- 
ter of 
Pin  in 
Inches. 

Area  of 
Pin  in 
Square 
Inches. 

foments  in  Inch-Pounds  for  Fiber  Strains  of 

Diame- 
ter of 
Pin  in 
Inches. 

15,000  Ib. 
>er  Square 
Inch. 

20,000  Ib. 
>er  Square 
Inch. 

22,000  Ib. 
aer  Square 
Inch. 

25,000  Ib. 
)er  Square 
Inch. 

1 

0.785 

1,470 

1,960 

2,160 

2,450 

1 

H 

0.994 

2,100 

2,800 

3,080 

3,500 

11 

H 

1.227 

2,880 

3,830 

4,220 

4,790 

H 

if 

1.485 

3,830 

5,100 

5,620 

6,380 

if 

H 

1.767 

4,970 

6,630 

7,290 

8,280 

H 

If 

2.074 

6,320 

8,430 

9,270 

10,500 

If 

If 

2.405 

7,890 

10,500 

11,570 

13,200 

if 

H 

2.761 

9,710 

12,900 

14,240 

16,200 

H 

2 

3.142 

11,800 

15,700 

17,280 

19,600 

2 

21 

3.547 

14,100 

18,800 

20,730 

23,600 

21 

2* 

3.976 

16,800 

22,400 

24,600 

28,000 

2^ 

2| 

4.430 

19,700 

26,300 

28,900 

32,900 

2f 

21 

4.909 

23,000 

30,700 

33,700 

38,400 

2| 

05 

ATC 

5.412 

26,600 

35,500 

39,000 

44,400 

2f 

o 

2f 

5.940 

30,600 

40,800 

44,900 

51,000 

2f 

21 

6.492 

35,000 

46,700 

51,300 

58,300 

21 

3 

7.069 

39,800 

53,000 

58,300 

66,300 

3 

31 

7.670 

44,900 

59,900 

65,900 

74,900 

31 

31 

8.296 

50,600 

67,400 

74,100 

84,300 

3i 

3| 

8.946 

56,600 

75,500 

83,000 

94,400 

3| 

9.621 

63,100 

84,200 

92,600 

105,200 

31 

sf 

10.321 

70,100 

93,500 

102,900 

116,900 

3f 

o 

3f 

11.045 

77,700 

103,500 

113,900 

129,400 

3f 

31 

11.793 

85,700 

114,200 

125,600 

142,800 

31 

4 

12.566 

94,200 

125,700 

138,200 

157,100 

4 

41 

13.364 

103,400 

137,800 

151,600 

172,300 

41 

o 

4 

14.186 

113,000 

150,700 

165,800 

188,400 

4^ 

4 

15.033 

123,300 

164,400 

180,800 

205,500 

4f 

44 

15.904 

134,200 

178,900 

196,800 

223,700 

41 

1 

16.800 

145,700 

194,300 

213,700 

242,800 

4£ 

4f 

17.721 

157,800 

210,400 

231,500 

263,000 

4f 

41 

18.665 

170,600 

227,500 

250,200 

284,400 

41 

150 


ARCHITECTURAL  ENGINEERING. 


RESISTING   MOMENTS    OF   PINS—  Continued. 


Diame- 
ter of 
Pin  in 
Inches. 

Area  of 
Pin  in 
Square 
Inches. 

Moments  in  Inch-Pounds  for  Fiber  Strains  o: 

Diame- 
ter of 
Pin  in 
Inches. 

15,000  Ib. 
per  Square 
Inch. 

20,000  Ib. 
per  Square 
Inch. 

22,000  Ib. 
per  Square 
Inch. 

25,000  Ib. 
per  Square 
Inch. 

5 

19.635 

184,100 

245,400 

270,000 

306,800 

5 

51 

20.629 

198,200 

264,300 

290,700 

330,400 

51 

H 

21.648 

213,100 

.   284,100 

312,500 

355,200 

4 

H 

22.691 

228,700 

304,900 

335,400 

381,100 

4 

H 

23.758 

245,000 

326,700 

359,300 

408,300 

51 

5| 

24.850 

262,100 

349,500 

384,400 

436,800 

5| 

5| 

25.967 

280,000 

373,300 

410,600 

466,600 

5f 

51 

27.109 

298,600 

398,200 

438,000 

497,700 

51 

6 

28.274 

318,100 

424,100 

466,500 

530,200 

6 

61 

29.465 

338,400 

451,200 

496,300 

564,000 

61 

H 

30.680 

359,500 

479,400 

527,300 

599,200 

4 

6f 

31.919 

381,500 

508,700 

559,600 

635,900 

4 

61 

33.183 

404,400 

539,200 

593,100 

674,000 

61 

6| 

34.472 

428,200 

570,900 

628,000 

713,700 

6f 

6| 

35.785 

452,900 

603,900 

664,200 

754,800 

4 

H 

37.122 

478,500 

638,000 

701,800 

797,500 

61 

7 

38.485 

505,100 

673,500 

740,800 

841,900 

7 

71 

39.871 

532,700 

710,200 

781,200 

887,800 

71 

7i 

41.282 

561,200 

748,200 

823,000 

935,300 

7i 

7f 

42.718 

590,700 

787,600 

866,300 

984,500 

7f 

71 

44.179 

621,300 

828,400 

911,200 

1,035,400 

71 

7f 

45.664 

652,900 

870,500 

957,500 

1,088,100 

7| 

7f 

47.173 

685,500 

914,000 

1,005,300 

1,142,500 

7f 

71 

48.707 

719,200 

958,900 

1,054,800 

1,198,700 

71 

8 

50.265 

754,000 

1,005,300 

1,105,800 

1,256,600 

8 

81 

51.849 

789,900 

1,053,200 

1,158,500 

1,316,500 

81 

H 

53.456 

826,900 

1,102,500 

1,212,800 

1,378,200 

8f 

8f 

55.088 

865,100 

1,153,400 

1,268,800 

1,441,800 

4 

81 

56.745 

904,400 

1,205,800 

1,326,400 

1,507,300 

81 

8f 

58.426 

944,900 

1,259,800 

1,385,800 

1,574,800 

4 

8f 

60.132 

986,500 

1,315,400 

1,446,900 

1,644,200 

4 

81 

61.862 

1,029,400 

1,372,500 

1,509,800 

1,715,700 

81 

ARCHITECTURAL  ENGINEERING. 


151 


RESISTING    MOMENTS    OF  PINS—  Continued. 


Moments  in  Inch-Pounds  for  Fiber  Strains  of 

Diame- 

Area of 

Diame- 

ter of 

Pin  in 

ter  of 

Pin  in 
Inches. 

Square 
Inches. 

15,000  Ib. 
per  Square 

20,000  Ib. 
per  Square 

22,000  Ib. 
per  Square 

25,000  Ib. 
per  Square 

Pin  in 
Inches. 

Inch. 

Inch. 

Inch. 

Inch. 

9 

63.617 

1,073,500 

1,431,400 

1,574,500 

1,789,200 

9 

H 

65.397 

1,118,900 

1,491,900 

1,641,100 

1,864,800 

9-1- 

91 

67.201 

1,165,500 

1,554,000 

1,709,400 

1,942,500 

91 

9| 

69.029 

1,213,400 

1,617,900 

1,779,600 

2,022,300 

9— 

9t 

70.882 

1,262,600 

1,683,400 

1,851,800 

2,104,300 

4 

72.760 

1,313,100 

1,750,800 

1,925,900 

2,188,500 

9| 

Of 

74.662 

1,364,900 

1,819,900 

2,001^900 

2,274,900 

9| 

91 

76.590 

1,418,100 

1,890,800 

2,079,900 

2,363,500 

91 

10 

78.540 

1,472,600 

1,963,500 

2,159,900 

2,454,400 

10 

101 

82.520 

1,585,900 

2,114,500 

2,325,900 

2,643,100 

101 

10| 

86.590 

1,704,700 

2,273,000 

2,500,200 

2,841,200 

104- 

lOf 

90.760 

1,829,400 

2,439,300 

2,683,200 

3,049,100 

10-1 

11 

95.030 

1,960,100 

2,613,400 

2,874,800 

3,266,800 

11 

111 

99.400 

2,096,800 

2,795,700 

3,075,400 

3,494,800 

111 

103.870 

2,239,700 

2,986,300 

3,284,800 

3,732,800 

1H 

12T 

113.100 

2,544,700 

3,392,900 

3,732,200 

4,241,200 

12 

153  ARCHITECTURAL    ENGINEERING. 

DEFLECTION    OF    BEAMS. 


Description. 


Mode  of  Loading. 
Lengths  in  Inches.  Loads  in  Pound 


Greatest  Deflection 
in  Inches. 


One  end  firm- 
ly fixed,  other 
end  loaded. 


Supported  at 
both  ends,  load- 
ed at  the  center. 


Supported  at 
both  ends,  load- 
ed any  place. 

One  end  fixed, 
other  end  sup- 
ported, loaded 
at  center. 

Both  ends 
fixed,  loaded  at 
center. 

Loaded  at 
each  end,  two 
supports  be- 
tween ends. 

Both  ends 
supported,  two 
symmetrical 
loads. 

One  end  fixed, 
load  uniformly 
distributed. 

Both  ends 
supported,  load 
uniformly  dis- 
tributed. 


WL3 
3ES 


WL* 


IVab  \/Za(2L  —  a)* 


W  LEI 

3  WL* 
322  El 

WL3 
192  El 

For  overhang: 


Between  supports: 


8ES 


384  El 


G  ARCHITECTURAL    ENGINEERING. 

IXEFJVECTIOX    OF    BKAMS—  Continued. 


153 


Description. 


Mode  of  Loading, 
^engths  in  Inches.  Loads  in  Pounds, 


Greatest  Deflection 
in  Inches. 


Both  ends 
fixed,  load 
uniformly  dis- 
tributed. 

One  end  fixed, 
load  distribu- 
ted, increasing 
uniformly  to- 
wards the  fixed 
ends,. 

Both  ends 
supported,  load 
distributed,  in- 
creasing u  n  i- 
formly  toward 
the  center. 

Both  ends 
supported,  loac 
distributed,  de- 
creasing u  n  i- 
formly  towards 
the  center. 

Both  end? 
supported,  loac 
increasing  uni 
formly  towards 
one  end. 


L 


384  A' 


YoEI 


\\' I* 

WEI 


3  Il'L3 


47  U'L3 
3,600  E I 


A  SERIES 


OF 


QUESTIONS  AND  EXAMPLES 

RELATING  TO  THE  SUBJECTS 
TREATED  OF  IN  THIS  VOLUME. 


It  will  be  noticed  that  the  various  Question  Papers  in  this 
volume  are  numbered  to  correspond  with  the  sections  to 
which  they  refer,  the  section  numbers  being  placed  on  the 
headline  opposite  the  page  number,  as  in  the  preceding 
sections.  As  in  the  case  of  the  Instruction  Papers,  each  sec- 
tion is  complete  in  itself,  the  page  numbers  and  question 
numbers  beginning  with  (1)  for  each  section. 


ARITHMETIC. 

(ARTS.  1-181.      SEC.  1.) 


(1)  What  is  arithmetic  ? 

(2)  What  is  a  number  ? 

(3)  What  is  the  difference  between  a  concrete  number 
and  an  abstract  number  ? 

(4)  Define  notation  and  numeration. 

(5)  Write  each  of  the  following  numbers  in  words : 

(a)  980  ;  (b)  605  ;  (c)  28,284  ;  (d)  9,000,042  ;  (c)  850,- 
317,002;  (/)  700,004. 

(6)  Represent  in  figures  the  following  expressions : 

(a)  Seven  thousand  six  hundred.  (/>)  Eighty-one  thousand 
four  hundred  two.  (c}  Five  million  four  thousand  seven. 
(d)  One  hundred  eight  million  ten  thousand  one.  (c)  Eight- 
een million  six.  (/)  Thirty  thousand  ten. 

(7)  What  is  the   sum   of   3,290  +  504  +  805,403  +  2,074 
+  81  +  7?  Ans.   871,359. 

(8)  709  +  8,304,725  +  391  + 100,302  +  300  +  909  =  what  ? 

Ans.   8,407,330. 

(9)  Find  the  difference  between  the  following: 

(a)  50,962  and  3,338;  (b)  10,001  and  15,339. 

I  (a)     47,624. 

I  (£)     5,338. 


2  ARITHMETIC.  §  1 

(10)  (a)  70,968-32,975  =  ?     (b)  100,000-98,735  =  ? 

(  (a)     37,993. 
Ans>  (•(£)     1,265. 

(11)  The  greater  of  two  numbers  is  1,004  and  their  differ- 
ence is  49;  what  is  their  sum  ?  Ans.   1,959. 

(12)  From  5,  962  +  8,  471  +  9,  023  take  3,  874  +  2,  039. 

Ans.   17,543. 

(13)  A  man  willed  $125,000  to  his  wife  and  two  children; 
to  his  son  he  gave  $44,675,  to  his  daughter  $26,380,  and  to 
his  wife  the  remainder.     What  was  his  wife's  share  ? 

Ans.   $53,945. 

(14)  Find  the  products  of  the  following: 

(a)  526,387X7;  (b)  700,298x17;  (c)  217  X  103  X  67. 

(  (a)     3,684,709. 
Ans.  •<  (t)     11,905,066. 
(  (c)     1,497,517. 

(15)  If  your  watch  ticks  once  every  second,  how  many 
times  will  it  tick  in  one  week  ?  Ans.   604,800  times. 

(16)  If  a  monthly  publication  contains  24  pages  in  each 
issue,  how  many  pages  will  there  be  in  8  yearly  volumes  ? 

Ans.  2,304. 

(17)  An  engine  and  boiler  in  a  manufactory  are  worth 
$3,246.     The  building  is  worth  three  times  as  much,  plus 
$1,200,  and  the  tools  are  worth  twice  as  much  as  the  build- 
ing, plus  $1,875.     (a)  What  is  the  value  of  the  building  and 
tools  ?     (b}  What  is  the  value  of  the  whole  plant  ? 

.         ((a)     $34,689. 
"•](*)     $37,935. 

(18)  Solve  the  following  by  cancelation: 
72X48X28X5  .,.   80x60x50x16x14 


96X15X7X6  v  '       70X50X24X20 

I 


A         .  («)     8" 
Ans<  1  (b)     32. 


§  1  ARITHMETIC.  3 

(19)  If  a  mechanic  earns  $1,500  a  year  for  his  labor,  and 
his  expenses  are  1908  per  year,  in  what  time  can  he  save 
enough  to  buy  28  acres  of  land  at  $133  an  acre  ? 

Ans.  7  yr. 

(20)  A  freight  train  ran  3G5  miles  in  one  week,  and  3 
times  as  far,  lacking  246  miles,  the  next  week ;  how  far  did 
it  run  the  second  week  ?  Ans.  849  mi. 

(21)  If  the  driving  wheel  of  a  locomotive  is  1C  feet  in 
circumference,  how  many  revolutions  will  it  make  in  going 
from  Philadelphia  to  Pittsburg,  the  distance  between  which 
is  354  miles,  there  being  5,280  feet  in  one  mile  ? 

Ans.   110,820  rev. 

(22)  What  is  the  quotient  of: 

(a)  589, 824 -f- 576?     (/;)   369,730,620-^-43,911?     (c)  2,527,- 
525-^505?     (d)  4,961,794,302-^1,23.4?         f  ^      i)024. 

(b]      8,420. 
Ans. -I  )  !        ' 

5,005. 

4,020,903. 

(23)  A  man  paid  $444  for  a  horse,  wagon,  and  harness. 
If  the  horse  cost  $264  and  the  wagon  $153,  how  much  did 
the  harness  cost  ?  Ans.  $27. 

(24)  What  is  the  product  of: 

(a)  1,024X576?     (£)  5,005x505?     (c)  43,911x8,420? 

(  (a]     589,824. 
Ans.  J  (£)     2,527,525. 
(  (c)     369,730,620. 

(25)  If  a  man  receives  30  cents  an  hour  for  his  wages, 
how  much  will  he  earn  in  a  year,  working  10  hours  a  day 
and  averaging  25  days  per  month  ?  Ans.  $900. 

(26)  What  is  a  fraction  ? 

(27)  What  are  the  terms  of  a  fraction  ? 

(28)  What  does  the  denominator  show  ? 

(29)  What  does  the  numerator  show  ? 

(30)  How  do  you  find  the  value  of  a  fraction  ? 


4  ARITHMETIC.  §  1 

(31)  Is  -1/  a  proper  or  an  improper  fraction,  and  why  ? 

(32)  Write  three  mixed  numbers. 

(33)  Reduce  the  following  fractions  to  their  lowest  terms : 
I,  T4e>  A.  ft-  Ans-  i»  i»  i.  *• 

(34)  Reduce  6  to  an  improper  fraction  whose  denomina- 
tor is  4.  Ans.  -2/. 

(35)  Reduce  7$,  13T\,  and  lOf  to  improper  fractions. 

63     218.     43 


(36)     What  is  the  value  of  each  of  the  following:  -1/,  -^, 
b  ~s~i  ¥¥  •  Ans.  OTJ-,  4^,  4y^-,  2, 


(37)     Solve  the  following: 

(a)     35-^;      (*)     ^-f-3;      (c)     ¥  +  *'>     (d)    W^A; 


(e)   15|H-4f. 


Ans.  • 


(a)     112. 
(*)     T3«r- 
W      «- 


(38)  l  +  f+.f  =  ?  Ans.  1. 

(39)  i  +  t+A  =  ?  Ans.  |f. 

(40)  42  +  31f  +  9^  =  ?  Ans.  83^. 


(41)  An  iron  plate  is  divided  into  four  sections;   the  first 
contains  29f  square  inches;  the  second,  50f  square  inches; 
the   third,  41  square  inches;    and  the  fourth,  69^-   square 
inches.     How  many  square  inches  are  in  the  plate  ? 

Ans.  190T93-  sq.  in. 

(42)  Find  the  value  of  each  of  the  following : 

15  4  +  3 

,7          ,x  32          ,2  +  6  ((a) 


Ans. 

'     6  '     D  i) 

16  8 

(43)     The  numerator  of  a  fraction  is  28,  and  the  value  of 
the  fraction  | ;  what  is  the  denominator  ?  Ans.  32. 


§  1  ARITHMETIC.  5 

(44)  What  is  the  difference  between  (<?)  £  and  ^  ?  (/>)  13 
and  7TV  ?    (c)  312^  and  22!)^  ? 

(  (a)     TV 
Ans.  .1  (b)     5TV 
1(0     83if. 

(45)  If  a  man  travels  85yV  miles  in  one  day,  78T9-  miles  in 
another  day,  and  125i|  miles  in  another  day,  how  far  did  he 
travel  in  the  three  days  ? 

Ans.  289  2-1- J-  mi. 

(46)  From  5734  tons  take  21  Of  tons.  Ans.  357^  T. 

(47)  At  f  of  a  dollar  a  yard,  what  will  be  the  cost  of  9^ 
yards  of  cloth  ?  Ans.  3||  dollars. 

(48)  Multiply  f  of  f  of  T7T  of  ia  of  11  by  £  of  f  of  45. 

Ans.  lofl^V 

(49)  How  many  times  are  f  contained  in  f  of  10  ? 

Ans.  18  times. 

(50)  Bought  21 1£  pounds  of  old  lead  for   1£  cents  per 
pound.      Sold  a  part  of  it  for  2|  cents  per  pound,  receiving 
for  it  the  same  amount  as  I  paid  for  the  whole.     How  many 

pounds  did  I  have  left  ? 

Ans.  52f|  Ib. 

(51)  Write   out  in   words  the   following  numbers:  .08, 
.131,  .0001,  .000027,  .0108,  and  93.0101. 

(52)  How  do  you  place  decimals  for  addition  and  sub- 
traction ? 

(53)  Give  a  rule  for  multiplication  of  decimals. 

(54)  Give  a  rule  for  division  of  decimals. 

(55)  State  the  difference  between  a  fraction  and  a  decimal. 

(56)  State  how  to  reduce  a  fraction  to  a  decimal. 

1-32 


ARITHMETIC. 


hials:  ^,  £,  -£%,  T%5T5->  and  T] 

Ans.  - 


(57)  Reduce  the  following-  fractions  to  equivalent  deci- 

5. 

875. 
15625. 
65. 

125. 

(58)  Solve  the  following: 

32.5  +  . 29+1.5.  (,}  1.283XET+5. 

~       4. 7   '  ° 


/  \   589  +  27X163-8,  40.6  +  7.1  X  (3.029-1.874) 

25  +  39  6.27  +  8.53-8.01 


Ans. 


(a)  2.5029. 

(b)  6.3418. 


(c)  1,491.875. 

(d)  8.1139. 

(59)  How  many  inches  in  .875  of  a  foot  ?        Ans.  10^  in. 

(60)  What  decimal  part  of  a  foot  is  T\  of  an  inch  ? 

Ans.  .015625. 

(61)  A  cubic  inch  of  water  weighs  .03617  of  a  pound. 
What  is  the  weight  of  a  body  of  water  whose  volume  is 
1,500  cubic  inches?  Ans.   54.255  Ib. 

(62)  If  by  selling  a  carload  of  coal  for  $82.50,  at  a  profit 
of  $1.65  per  ton,  I  make  enough  to  pay  for  72.6  feet  of 
fencing  at  $.50  a  foot,  how  many  tons  of  coal  were  in  the 
car?  Ans.  22  T. 

(63)  Divide  17,892  by  231,  and  carry  the  result  to  four 
decimal  places.  Ans.   77.4545+. 

(64)  What  is  the  value  of  the  following  expression  car- 
ried to  three  decimal  places: 

74.  26  X  24  X  3.  1416  X  19  X  19  X  350 


33,000X12X4 


(65)     Express:     (a)  .7928  in  64ths;     (b)  .1416   in   32ds; 
(c)  .47915  in  16ths.  r  (a)     ££. 

Ans.     (b)     &. 


ARITHMETIC. 


Ans. 


(«) 

w 


(GG)     Work  out  the  following  examples: 
(a)    709.03-. 8514;     (/;)    81.903-1.7;     (c)    18-. 18;    (d) 
1-.001;     (c)     872.1 -(.8721 +  .008);      (/)     (5.028  +  .0073) 
-(6.704-2.38).  |-  (a)     708.7786. 

80.203. 
17.82. 
.999. 

871.2199. 
.7113. 

(67)     Work  out  the  following: 

(a)  $-.807;  (/>)  .875 -f;  (c)  (^  +  .435) -(TVy-.07); 
(d]  What  is  the  difference  between  the  sum  of  33  millionths 
and  17  thousandths,  and  the  sum  of  53  hundredths  and  274 
thousandths?  (#)  .068. 

(//)      .5. 

(c)  .45125. 

(d)  .786907. 

(08)  What  is  the  sum  of  .125,  .7,  .089,  .4005,  .9,  and 
.000027?  Ans.  2.214527. 

(69)  927.410  +  8.274  +  372.0  +  02.07938  =  ? 

Ans.    1,370.30938. 

(70)  Add  17  thousandths,  2  tenths,  and  47  millionths. 

Ans.   .217047. 

(71)  Find  the  products  of  the  following  expressions: 

(a)  .013X.107;  (/;)  203  X  2.03  X  .203;  (c)  2.7x31.85  X  (3. 10 
—  .310);  (d)  (107.8  +  0. 541-31. 90)xl. 742. 

(«) 
(*) 


Ans. 


(c) 
(d) 


.001391. 
83.05427. 
244.50978. 
143.507702. 


(72)     Solve  the  following: 
(a)  (A-.13)X.625  +  |; 
.013-2.17)Xl3i-7fV 


X  .21)  -  (.02x  A);  (c)  (-'/ 

f  (<0     -384375. 
Ans.  J  (/;)     .1209375. 

I  (c)     6.4896875. 


ARITHMETIC. 


(73)     Solve  the  following  : 
(a)  .875  -H;  (*)  i--55 


Ans.  < 


\  (a)     1.75. 
(b)     1.75. 


(74)     Find  the  value  of  the  following  expression : 
1.25X20X3 


87  +  (11X8)' 
459  +  32 


Ans.  21  Of 


(75)     From  1  plus  .001  take  .01  plus  .000001. 

Ans.  .990999. 


ARITHMETIC. 

(ARTS.  1-168.     SEC.  2.) 


(1)  What  is  25  per  cent,  of  8,428  lb.?  Ans.  2,1071b. 

(2)  What  is  1  per  cent,  of  $100  ?  Ans.  $1. 

(3)  What  is  %  per  cent,  of  $35,000  ?  Ans.   $175. 

(4)  What  per  cent,  of  50  is  2  ?  Ans.  -if. 

(5)  What  per  cent,  of  10  is  10  ?  Ans.  100#. 

(6)  Solve  the  following: 

(a)  Base  =  $2,522  and  percentage  =  $176.54.  What  is 
the  rate?  (b]  Percentage  =  16.96  and  rate  =  8  per  cent. 
What  is  the  base  ?  (c)  Amount  =  2  16.  7025  and  base  =  213.5. 
What  is  the  rate?  (d)  Difference  =  201.825  and  base 
=  207.  What  is  the  rate  ? 

212. 


Ans. 


(c) 


(7)  A  farmer  gained   15$  on  his  farm  by  selling  it  for 
$5,500.     What  did  it  cost  him  ?  Ans.  $4,782.61. 

(8)  A  man  receives  a  salary  of  $950.      He  pays  24^  of  it 
for  board,  12^  of  it  for  clothing,  and  17#  of  it  for  other 
expenses.     How  much  does  he  save  in  a  year  ?    Ans.  $441.75. 

(9)  If  37£  per  cent,  of  a  number  is  961.38,  what  is  the 
number?  Ans.   2,563.68. 


2  ARITHMETIC.  §  2 

(10)  A  man  owns  f  of  a  property.     30$  of  his  share  is 
worth  $1,125.     What  is  the  whole  property  worth  ? 

Ans.  $5,000. 

(11)  What  sum  diminished  by  35$  of  itself  equals  $4,810  ? 

Ans.  $7,400. 

(12)  A  merchant's  sales  amounted  to  $197.55  on  Monday, 
and  this  sum  was  12£$  of  his  sales  for  the  week.    How  much 
were  his  sales  for  the  week  ?  Ans.  $1,580.40. 

(13)  The  distance  between  two  stations  on  a  certain  rail- 
road is  1G.5  miles,  which  is  12£$  of  the  entire  length  of  the 
road.    What  is  the  length  of  the  road  ?  Ans.   132  mi. 

(14)  After  paying  60$  of  my  debts  I  find  that  I  still  owe 
$35.     What  was  my  whole  indebtedness  ?  Ans.  $87.50. 

(15)  Reduce  28  rd.  4  yd.  2  ft.  10  in.  to  inches. 

Ans.  5, 722  in. 

(1C)     Reduce  5,722  in.  to  higher  denominations. 

Ans.  28  rd.  4  yd.  2  ft.  10  in. 

(17)  How  many  seconds  in  5  weeks  and  3.5  days  ? 

Ans.  3,326,400  sec. 

(18)  How   many    pounds,    ounces,     pennyweights,    and 
grains  are  contained  in  13,750  gr.  ? 

Ans.  2  Ib.  4  oz.  12  pwt.  22  gr. 

(19)  Reduce  4,763,254  links  to  miles. 

Ans.  595  mi.  32  ch.  54  li. 

(20)  Reduce  764,325  cu.in.  to  cu.yd. 

Ans.   16  cu.yd.  10  cu.ft.  549  cu.in. 

(21)  What  is  the  sum  of  2  rd.  2  yd.  2  ft.  3  in. ;  4  yd.  1  ft. 
9  in. ;  2  ft.  7  in.  ?  Ans.  3  rd.  2  yd.  2  ft.  1  in. 

(22)  What   is  the  sum  of  3  gal.  3  qt.  1  pt.  3  gi. ;    6  gal. 
1  pt.  2  gi. ;  4  gal.  1  gi. ;  8  qt.  5  pt.  ?     Ans.   16  gal.  3  qt.  2  gi. 

(23)  What  is  the  sum  of  240  gr.  125  pwt.  50  oz.  and  3  Ib.  ? 

Ans.  7  Ib.  8  oz.  15  pwt. 


§2  ARITHMETIC.  3 

(24)  What  is  the  sum  of  11°   10'  12";    13°  19' 30";  20° 
25";  26'  29";  10°  17'  11"  ?  Ans.  55°  19'  47". 

(25)  What  is  the  sum  of  130  rd.  5  yd.  1  ft.  6  in. ;  215  rd. 

2  ft.  8  in. ;  304  rd.  4  yd.  11  in.  ?    Ans.   2  mi.  10  rd.  5  yd.  7  in. 

(26)  What  is  the  sum  of  21  A.  67  sq.ch.  3  sq.rd.  21  sq.li. ; 
28  A.  78  sq.ch.  2  sq.rd.  23  sq.li.;  47  A.  6  sq.ch.  2  sq.rd. 
18  sq.li.  ;  56  A.  59  sq.ch.  2  sq.rd.  16  sq.li.  ;  25  A.  38  sq.ch. 

3  sq.rd.  23  sq.li.  ;  46  A.  75  sq.ch.  2  sq.rd.  21  sq.li.? 

Ans.  255  A.  3  sq.ch.  14  sq.rd.  122  sq.li. 

(27)  From  20  rd.  2  yd.  2  ft.  9  in.  take  300  ft. 

Ans.  2  rd.  1  yd.  2  ft.  9  in. 

(28)  From  a  farm  containing  114  A.  80  sq.rd.  25  sq.yd., 
75  A.  70  sq.rd.  30  sq.yd.  are  sold.      How  much  remains  ? 

Ans.   39  A.  9  sq.rd.  25^  sq.yd. 

(29)  From  a  hogshead  of  molasses,  10  gal.  2  qt.  1  pt.  are 
sold  at  one  time,  and  26  gal.  3  qt.  at  another  time.     How 
much  remains  ?  Ans.  25  gal.  2  qt.  1  pt. 

(30)  If  a  person  were  born  June  19,  1850,  how  old  would 
he  be  August  3,  1892  ?  Ans.  42  yr.  1  mo.  14  da. 

(31)  A  note  was  given  August  5,    1890,    and   was   paid 
June  3,  1892.     What  length  of  time  did  it  run  ? 

Ans.   1  yr.  9  mo.  28  da. 

(32)  What  length  of    time    elapsed  from  16  min.   past 
10  o'clock  A.  M.,  July  4,  1883,  to  22  min.  before  8  o'clock  P.  M., 
Dec.  12,  1888  ?  Ans.  5  yr.  5  mo.  8  da.  9  hr.  22  min. 

(33)  If  1  iron  rail  is  17  ft.  3  in.  long,  how  long  would 
51  rails  be,  if  placed  end  to  end  ?         Ans.  53  rd.  1^  yd.  9  in. 

(34)  Multiply  3  qt.  1  pt.  3  gi.  by  4.7. 

Ans.  4  gal.  2  qt.  1.7  gi. 

(35)  Multiply  3  Ib.  10  oz.  13  pwt.  12  gr.  by  1.5. 

Ans.  5  Ib.  10  oz.  6  gr. 


4  ARITHMETIC.  §  2 

(36)  How  many  bushels  of  apples  are  contained  in  9  bbl., 
if  each  barrel  contains  2  bu.  3  pk.  6  qt.  ? 

Ans.   26  bu.  1  pk.  6  qt. 

(37)  Multiply  7  T.  15  cwt.  10.5  Ib.  by  1.7. 

Ans.   13  T.  3  cwt.  67.85  Ib. 

(38)  Divide  358  A.  57  sq.rd.  6  sq.yd.  2  sq.ft.  by  7. 

Ans.   51  A.  31  sq.rd.  8  sq.ft. 

(39)  Divide  282  bu.  3  pk.  1  qt.  1  pt.  by  12. 

Ans.   23  bu.  2  pk.  2  qt.  £  pt. 

(40)  How  many  iron  rails,  each  30  ft.  long,  are  required 
to  lay  a  railroad  track  23  mi.  long  ?  Ans.  8,096  rails. 

(41)  How  many  boxes,   each  holding  1  bu.  1  pk.  7  qt., 
can  be  filled  from  356  bu.  3  pk.  5  qt.  of  cranberries  ? 

Ans.  243  boxes. 

(42)  If  16  square  miles  are  equally  divided  into  62  farms, 
how  much  land  will  each  contain  ? 

Ans.   165  A.  25  sq.rd.  24  sq.yd.  3  sq.ft.  80+  sq.in. 

(43)  What  is.  the  square  of  108  ?  Ans.   11,664. 

(44)  What  is  the  cube  of  181.25  ?  Ans.  5,954,345.703125. 

(45)  What  is  the  fourth  power  of  27.61  ? 

Ans.  581,119.73780641. 

(46)  Solve  the  following:  (#)1062;  (b}  (182|)2;  (c}  .005'; 
(d)  .0063';  (e)  10. 062.  f  (a)      11,236. 

(b)  33,169.515625. 

Ans.  -    (c)  .000025. 

(d)  .00003969. 

(e)  101.2036. 

(47)  Solve  the  following:  (a)  753s;  (b)  987.4s;  (c)  .0053; 
(d]  .40443.  f  (^      426,957,777. 

(b)      962,674,279.624. 

J\  f*  C  »    * 

1  (c)   .000000125. 
(d\  .066135317184. 


§  2  ARITHMETIC. 

(48)  What  is  the  fifth  power  of  2  ? 

(49)  What  is  the  fourth  power  of  3  ? 


Ans.  32. 
Ans.   81. 


(50)     What  are  the  values  of:  (a)  67. 853?    (d)  967, 8453? 

to  d)'?  (d)  (ir? 

(a)  4,603.6225. 

(b)  936,723,944,025. 


Ans. 


to 


9 

6T- 


(51)  What  is  (a)  the  tenth  power  of  5  ?    (b)  the  fifth  power 
of  9?  j  (a)     9,765,625. 

'   (  (&)     59,049. 

(52)  Solve  the  following:  (a)  1.24;  (b)  II8;  (c)  I7;  (,/)  .01'; 


to  -r- 


Ans. 


(a)  2.0736. 

(b)  1,771,561. 

(c)  i. 

(^)  .00000001. 

^  .00001. 


(53)     Find  the  values   of  the  following:    (a)  .01333;    (b) 


301.011s;  (c)  (kY\  (d}  (3f)3. 

Ans. 


(a)  .000002352637. 

(6)  27,273,890.942264331. 

to  ^r- 

(d)  52^-,  or  52.734375. 


(54)  In  what  respect  does  evolution  differ  from  involution  ? 

NOTE. — In  the  answers  to  the  following  examples,  a  minus  sign  after 
a  number  indicates  that  the  last  digit  is  not  quite  as  large  as  the  num- 
ber printed.  Thus,  12.497—  indicates  that  the  number  really  is  12.41*64- , 
and  that  the  6  has  been  made  a  7  because  the  next  succeeding  figure 
was  5  or  greater.  For  example,  had  it  been  desired  to  use  but  three 
decimal  places  in  example  46  (b),  the  answer  would  have  been  written 
33,169.516-. 

(55)  Find  the  square  root  of  the  following:  (a)  3,486,- 
784.401;  (b)  9,000,099.4009;  (c)  .001225. 

(a)     1,867.29+. 


Ans. 


(b)     3,000.017-. 
I  (c)     .035. 


ARITHMETIC. 


(56)  Extract  the  square  root  of  (a)  10,  795.  21;  (b)  73,008.04; 
(c)  90;  (d)  .09.  f  (a)      103.9. 

(b)      270.2. 
*       (,)      9.487-. 
[(d)     .3. 

(57)  Extract  the  cube  root  of  (a)  .32768;  (b)  74,088;  (c) 
92,416;  (d)  .373248.  f    ^)      .6894+. 


Ans 
AnS' 


45.212-. 
.72. 


(58)  Extract  the  cube  root  of  2  to  six  decimal  places. 

Ans.  1.259921+  . 

(59)  Extract    the    cube    root    of    (a)    1,758.416743;    (b) 
1,191,016;  (c)  ^;  (d)  jfr.  f  (a)      12.07. 

b)      106. 

w   *. 
w  t- 


Ans- 


(60)     Extract  the  cube  root  of  3  to  six  decimal  places. 

Ans.   1.442250—. 


(61)     Solve  the  following:  (a)  V123.21;  (b)  4/114.921;  (c) 

(a)      11.1. 


'502,681;  (d)  V.  00041209. 


Ans. 


(b)  10.72+. 

(c)  709. 

(d)  .0203. 


(62)     Solve    the    following:    (a)  ^.0065;    (b)  4^.021;    (c) 
4^8,036,054,027;  (d)  4^.000004096;  (e)  4^17. 

(a)  .18663-. 

(b)  .2759-. 
Ans.  -{  (c)  2,003. 

(</)  .016. 

^  2. 5713- . 


2  ARITHMETIC.  7 

(63)     Solve  the  following:  (a)    f  6,561;  (t>)    I'll 7, 649;  (c) 


.  000064;   (d)   ff. 


Ans 
AnS' 


(«)  9. 

(/;)  7" 

(.)  .2. 

(d)  .72112+. 


(64)     Extract    the    square   root  of:    (a)   ^ff;    (/;)   .3364; 


(c)  .1;  (</)  25.0|;  (c)  .OOOf. 


Ans. 


(«)  ff- 

(/;)  .58. 

(c)  .31623-. 

(d)  5. 00749+ . 

(e)  .02108+. 


(65)     (a)  Extract   the   fourth  root  of  2  to  four   decimal 
places,      (b)  Extract  the  sixth  root  of  6. 

L1893+' 


Ans 
" 


\(d)     1.  34801-  . 


(66)  Extract  the  square  root  of  (a)  3.1416  and  (6)  .7854 
to  four  decimal  places.  .         ((a)     1.7725—. 

M(*)     .8862+. 

(67)  Extract  the  cube  root  of  (a)  3.1416  and  (V)  .5236  to 
four  decimal  places.  .         \  (a)     1.4646—. 

'•}(£)     .8060-. 

Find  the  value  of  x  in  the  following  : 

(68)  11.7:  13::  20:  jr.  Ans.  22.22+. 

(69)  (a)  20  +  7:10  +  8::3:;r;  (b)  122  :  1002  ::  4  \x. 


Ans      (a 


10          x 

150  ~  600* 


Ans.  - 


\(d)     277.7+. 


(«)  x  =  12. 

(^)  jr  =  12. 

(c)  x  =  20. 

(d)  .1-  =  180. 

(e)  x  =  40. 


8                                     ARITHMETIC.  §  2 

(71)  .r:  5::  27: 12. 5.  Ans.  10$. 

(72)  45:  60::*:  24.  Ans.  18. 

(73)  *:35::4:7.  Ans.  20. 

(74)  9:*::  6: 24.  Ans.  36. 

(75)  ^1,000:^1,331  =  27:*.  Ans.  29.7. 

(76)  64:81  =  21*  :*2.  Ans.  23.625. 

(77)  7  +  8  :  7  =  30 :  x.  Ans.   14. 

(78)  A  man  whose  steps  measure  2  ft.  5  in.  takes  2,480 
steps  in  walking  a  certain  distance.     How  many  steps  of  2  ft. 
7  in.  will  be  required  for  the  same  distance  ? 

Ans.  2,320  steps. 

(79)  If  a  horse  travels  12  mi.  in  1  hr.  36  min.,  how  far 
will  he  travel  at  the  same  rate  in  15  hr.  ?          Ans.   112.5  mi. 

(80)  If  a  column  of  mercury  27.63  in.  high  weighs  .76  of 
a  pound,  what  will  be  the  weight  of  a  column  of  mercury 
having  the  same  diameter,  29.4  in.  high  ?         Ans.  .808+  Ib. 

(81)  If  2  gal.  3  qt.  1  pt.  of  water  will  last  a  man  5  da.,  how 
long  will  5  gal.  3  qt.  last  him,  if  he  drinks  at  the  same  rate  ? 

Ans.   10  da. 

(82)  Heat  from  a  burning  body  varies  inversely  as  the 
square  of  the  distance  from  it.     If  a  thermometer  held  6  ft. 
from  a  stove  shows  a  rise  in  temperature  of  24°,  how  many 
degrees  rise  in  temperature  would  it  indicate  if  held  12  ft. 
from  the  stove  ?  Ans.  6°. 

(83)  If  a  pile  of  wood  12  ft.  long,  4  ft.  wide,  and  3  ft.  high 
is  worth  $12,  what  is  the  value  of  a  pile  of  wood  15  ft.  long, 
5  ft.  wide,  and  6  ft.  high  ?  Ans.  $37.50. 

(84)  If  100  gal.  of  water  run  over  a  dam  in  2  hr.,  how 
many  gallons  will  run  over  the  dam  in  14  hr.  28  min.  ? 

Ans.  723£  gal. 


§  2  ARITHMETIC.  0 

(85)  If  a  cistern  28  ft.  long,  12  ft.  wide,  10  ft.  deep  holds 
510  .bbl.  of  water,  how  many  barrels  of  water  will  a  cistern 
hold  that  is  20  ft.  long-,  17  ft.  wide,  and  G  ft.  deep  ? 

Ans.  309T9T  bbl. 

(86)  If  a  railway  train  runs  444  mi.  in  8  hr.  40  min.,  in 
what  time  can  it  run  1,000  mi.  at  the  same  rate  of  speed  ? 

Ans.  20  hr.  41.44  min. 

(87)  If  sound  travels  at  the  rate  of  6,100  ft.  in  54  sec., 
how  far  does  it  travel  in  1  min.?  Ans.    07,200  ft. 

(88)  If  5  men  by  working  8  hours  a  day  can  do  a  certain 
amount  of  work,  how   many  men  by  working  10   hours   a 
day  can  do  the  same  work  ?  Ans.   4  men. 

(89)  If  a  man  travels  540  miles  in  20  days  of  10  hours 
each,  how  many  hours  a  day  must  he  travel  to  cover  0)50 
miles  in  25  days  ?  Ans.    9i  hr. 

(90)  Referring  to  example  4,  Art.  168,  Arithmetic,  g  2, 
what  is   the    horsepower  of   an   engine   whose   cylinder  is 
30   inches  in  diameter,  piston  speed,  000  feet  per  minute, 
and  mean  effective  pressure,  42  pounds  per  square  inch  ? 

Ans.   594  horsepower. . 

(91)  The    weight  of   a  cubic    inch   of  cast    iron   is  .201 
pound.     Referring   to  Art.  164,  Aritlimctic,   ^  2,  what  is 
the  weight  of  a  solid  cast-iron  cylinder  whose  diameter  is 
12  inches  and  length  is  60  inches  ?  Ans.   1,771.11  Ib. 

(92)  Referring   to  Art.    167,  Arithmetic,  §  2,   what   is 
the   centrifugal  force    of  a  40-pound  body  revolving  in   a 
circle  having  a  radius  of  10  inches,  at  a  speed  of  18  feet  per 
second  ?  Ans.  484. 7  Ib. 


FORMULAS. 

(ARTS.  1-21.     SEC.  3.) 


A  =  5  //  =  200 

B  =  10  x  =  12 

i  =  3.5  D  =  120 

Work  out  the  solutions  to  the  following  formulas,  using 
the  above  values  for  the  letters: 

(1)      C  =  Ans.    C  =  8. 

Ans"       = 


(3)      r  =  3-  246  B1t  .  Ans.  r  =  187.2G9+ 


Ans.  v  =  4. 05+. 
u  -   //_         "*A .  Ans.  ;/  =  G.35+. 

/w  f\n.f\-t  r*    7.  /     yt2  -,\ 

Ans.  /=  12,800. 


FORMULAS. 


Ans.  g  =  5. 


(8)      k  = 


Ans.  k  —  7.071+. 


(9)     T  = 


Ans.    T  =  10. 


GEOMETRY  AND  MENSURATION. 


(ARTS.   1-158.     SEC.  4.) 


(1)  Fig.  1  represents  a  center — half  a  regular  hexagon — for 
a  5-foot  semicircular  arch. 

(a)  What   is    the    angle   of 

bevel  (A  B  O)  to  which  the 

ends  of  the  planks  must  be 

cut?     (b}  What   length   of 

plank,  as  F D,  is  needed  for  / '\--.-t 

each  piece  ?   (c)  What  is  the 

length  of  the  inner  side,  as 

A  B,  of  each  plank,  after  cutting,  if  the  width  is  8  inches  ? 

t(a)     GO0. 
Ans.   !  (/;)     2.80  ft.  =  2  ft.  10.7  in. 

(  (c)     2.12  ft.  =  2  ft.  H  in. 

(2)  How  many  feet  of  molding  are  needed  to  go  around 

an  8-sided  room,  in  which 
the  distance  between  the 
parallel  sides  is  12  feet, 
and  between  the  opposite 

^T  ~n£-  y^   angles,  13  feet  ? 

Ans.  40  ft. 

ADB,  Fig.  2,  rep- 
an  arch  of  3  feet 
(a)  The  span  A  B 
feet,  what  is  the 


GEOMETRY  AND  MENSURATION. 


rise  CD1    (b)  If  the  radius  O  A,  and  rise   CD  were  given, 
show  how  to  find  A  B.  Ans.  (a)  1.02  ft.  =  1  ft.  \  in. 

(4)  In  staking  out  a  building,  the  points^, B,  C,D,E, 
and  F  are  located,  the  distances 
being  as  shown  in  Fig.  3.  To  check 
these  measurements,  it  is  desired  to 
measure  the  diagonals  FB,  FD,  EA, 
and  E  C.  What  are  their  lengths  ? 
Ans.  FB  =  30  ft. ;  FD  =  33.11  ft. ; 
EA  =  38.42  ft. ;  E  C  =  18.44  ft.  ' 

(5)     Explain  why  the  sides  of  a 
regular  hexagon  are   equal   to  the 
radius  of  the  circumscribing  circle. 

(6)  In  a  half-pitch  (or  45°)  roof  of  24-foot  span,  show  that 
the  length  along  the  roof  from  side  wall  to  ridge  =  i/288  feet. 

(7)  Show  by  a  sketch  how  to  divide  a  line  6  inches  long 
into  7  equal  parts. 

(8)  In   Fig.    4,    what    is    the 
length  of  the  'ridge  plate  A  B, 
which  is  10  feet  above  CD,  the 
latter  being  36  feet  long,  and  the  c. 


*ft*lf  bWA       M\*A.±*.fZ      U\J     J-\_>\_/  I    J.WXi41  •     CLJ.J.V4.      U1J.V    f*/  t*-S  T  1-* ATi 

angles  DC  A   and    CDS  being-   \*—  -36— 

45°?  Ans.   16  ft.  FlG-4- 

(9)  Fig.  5  represents  a 
roof  whose  span  A  B  is  22 
feet.  The  slopes  are  half- 
pitch,  and  the  distance  from 
E  to  D  (under  C)  is  11  feet. 

(a)  Find    the    length    of  a 
"common  rafter,"  as  G  H . 

(b)  Find    the    length   of   a 
" hip  rafter,"  as  B  C. 

(  (a)  15.56ft.  =  15  ft.  6|  in. 
FIG.  5.  (  (b}  19.05  ft.  =  19  ft.  f  in. 


Ans. 


GEOMETRY  AND  MENSURATION. 


(10)  In  Fig.  G,  the  distances 
from  A  to  C,  and  from  B  to  D, 
across  a  stream,  are  required.  The 
line  A  B  is  210  feet  long,  and  E  is 
the  middle  point.  E  F,  parallel  to 
B  D,  is  86  feet  long ;  and  E  G,  par- 
allel to  A  C,  is  90  feet  long.  What 
are  the  lengths  of  A  C  and  B  D  ? 


.A. 


7) 

••-Y 


Ans. 


f  ^  C,  180  ft. 
(  BD,  172ft. 


Ans. 


(11)  In  a  right  triangle,  one  of  the  acute  angles  is  37°. 
(a)  What  is  the  other  acute  angle  ?     (b)  What  is  the  angle 
formed  by  producing  one  side  of  the  given  angle  ? 

(a)     53'J. 
(/;)     143°. 

(12)  What  is  the  angle  included  between  two  adjacent 
sides  of  a  regular  nonagon  (nine-sided  polygon)  ? 

Ans.   140°. 

(13)  In  Fig.  7,  the  span  A  C  of  a  semicircular  arch  is  1G 

feet,  and  the  distance  A  E  to  the 
top  of  the  masonry  is  11  feet. 
How  thick  is  the  stonework  at 
D  F,  4  feet  from  the  vertical 
\ine  AEJ  Ans.  4.07  ft. 

(14)     Show  how  to  lay  out,  with 
a  tape,  a  line  12  feet  long,  perpen- 
dicular  to   another  at  its  middle 
point,  the  length  of  the  latter  being  32  feet. 

(15)  The  corners  of  a  cast-iron  plate,  15  inches  square 
and  1±  inches  thick,  are  rounded  off  by  quarter  circles  of 
2-inch  radius.     Estimating  cast  iron  at  .2G  pound  per  cubic 
inch,  how  much  less  does  this  plate  weigh  than  if  the  corners 
were  square  ?  Ans.  1.11  Ib. 

(16)  A  brick  pier  is  30  inches  square  at   the    base,  18 
inches  square  at  the  top,  and  6  feet  high.     Figuring  22$ 
brick  per  cubic  foot,  how  many  brick  are  required  for  18 
piers?  Ans.  9,922.5. 


GEOMETRY  AND  MENSURATION. 


(17)  Fig.  8  represents  a   cross-section   of  a    "Phoenix" 

column,  made  of  wrought  iron,  of 
which  a  strip  1  foot  long  and  1 
inch  square  in  section  weighs  3.33 
pounds.  Disregarding  curved  cor- 
ners and  edges,  but  adding  2  per 
cent,  to  the  weight  of  the  iron  for 
rivet  heads,  what  is  the  weight  of 
the  column  per  foot  of  length  ? 

Ans.  44.87  Ib. 

(18)  The  freight  rate  from  a  sandstone  quarry  to  a  town 
is  21  cents  per  hundred  pounds.     The  following  pieces  of 
dimension  stone  were  shipped.     Allowing  140  pounds  to  the 
cubic  foot,  what  were  the  charges  ? 

1  piece    4' 6"  X  3' 9"  X  15".     2  pieces  5'  6"  X  4'  0"x  15". 

1  piece    5'0"X3'9"X15". 

1  piece    4' 9"  X  4' 6"  X  15". 

2  pieces4/6"x4/3"Xl5". 

Ans.  $76.90. 

(19)  Fig.  9  represents  the  plan  and  end  views  of  a  roof. 

,  " 

10    J>  10* 


1  piece  4' 3"  X  3' 6"  X 15". 
1  piece  5/3'/x4/0//Xl5". 
3  pieces  3' 3"  X  3' 6"  X  15". 


Knowing  that  the  ridges  A  B  and  CD  are  10  feet  above  the 
eaves,  find  (a)  the  total  area  of  the  roof,  and  (b)  the  length 
of  the  «  valleys  "  F  C  and  E  C. 

.         I  (a)     1,300.88  sq.  ft. 

'  |  (6)     17.32  ft.  =  17  ft.  3|  in. 


GEOMETRY  AND  MENSURATION. 


(20)  Figuring    7-|-    gallons   per   cubic    foot,    how    many 
gallons  will  be  discharged  per  minute  through  a  4-inch  pipe, 
if  the  water  flows  5  feet  per  second  ?  Ans.  202.5  gal. 

(21)  What  length  of  lead  pipe,  weighing  .41  pound  per 
cubic  inch,  If  inches  outside  diameter  and  •£•  inch  thick,  will 
be  needed  to  make  a  3-pound  weight  ?  Ans.  11.6  in. 

(22)  The  span  A  B  of 
the   arch   shown  in  Fig. 
10  is  6  feet  8  inches,  and 
the  rise  CD  is  8  inches. 
(a)    Find  the  radius.     (/;) 
If  there  are  13  ring  stones, 
what  is  the  bottom  width 

of  each  ? 

(a)     8  ft.  8  in. 
£      G.32  in. 


Ans. 


(23)  In  Fig.  10,  find  the  FIG-  10- 

radius  by  the  principle  that  a  perpendicular  from  any  point 
on  a  circumference  to  a  diameter  is  a  mean  proportional 
between  the  two  parts  of  the  diameter. 

(24)  It  is  required,  in  a  question  of 
water  supply,  to  determine  the  area  of 
the  water  section  in  a  12-inch  pipe  flow- 
ing 9  inches  deep  ;  that  is,  what  is  the 
area  below  the  surface  A  />,  Fig.  11  ? 

Ans.  91.11  sq.  in. 

(25)  Find  the  total  feet  B.  M.  in  the 
following  bill  of  material  : 


FIG.  11. 


3  pieces  3"  X  10" 
8  pieces  3"  X  10" 
6  pieces  3"  X   8" 

10  pieces  3"  X   8" 

4  pieces.  3."  X   6" 
20  pieces  3"  X    6" 


;  1C'  0"  long. 
;  12'  6"  long. 
;  14'  6"  long. 
;  9'  0"  long. 
;  10'  6"  long. 
;  7'  0"  long. 


8  pieces  4"  XG";  18'  0"  long. 
4  pieces  4"  XG";    G'  6"  long. 
52  pieces  2"  X  4";  18'  0"  long. 
1  5  pieces  2  "  X  4  "  ;    9  '  G  "  long. 
GOO  ft.  B.  M.  1"  X  G"  flooring. 
200  ft.  B.  M.  2"  plank. 

Ans.  2,856  ft.  B.  M. 


GEOMETRY  AND  MENSURATION. 


12—  - 

1 

i 

i 

?>, 

Kt 

1 

<»^' 

FIG.  12. 


(26)  Fig,  12  represents  the  plan 
of  a  house  which  is  to  be  faced  with 
pressed  brick.  The  walls  are  22  ft. 
high,  and  there  are  two  gables,  one 
14  ft.  X  7  ft.  high,  and  the  other 
12  ft.  X  7  ft.  high.  Deducting  10 
windows,  3  ft.  X  7  ft. ;  10  windows, 
3  ft.  X  6  ft. ;  and  4  doors,  3  ft.  6  in. 
X  8  f  t. ;  and  allowing  7  brick  to  each 
square  foot  of  surface,  how  many 
brick  will  be  required  ?  Ans.  11,907. 

(27)  The  major  and  minor  axes  of  an  elliptic  window 
frame  are  8  feet  and 

5  feet,  respectively. 
The  sash  is  4  inches 
wide  all  around. 
What  is  the  area  of 
the  glass  ? 

Ans.   24.93  sq.ft. 

(28)  A  culvert  is 
required  to  have  an 


FIG.  i.s. 

area  of  88  square  feet.  How  much 
more  than  this  area  is  the  cross-sec- 
tion shown  in  Fig.  13,  the  arch  being 
semi-elliptical?  Ans.  3.42  sq.  ft. 

(29)  How  many  cubic  yards  are 
there  in  a  retaining  wall  16  feet  long, 
having    the    cross-section   shown  in 
Fig.  14?  Ans.  42.36  cu.  yd. 

(30)  The   top  of  the  frame  of    a 
window  3|  feet  in  width  is  to  be  bent 
to  a  circular  arc,  the  rise  of  which  is  3 
inches.     What  length  of  piece  will  be 
required  ? 

Ans.  42.57  in.  =  3  ft.  6^  in. 


GEOMETRY  AND  MENSURATION. 


(31)  Compute  the  cubic  yards  of  excavation  for  the  cellar, 
9  feet  deep,  of  the  building  shown  in  Fig.  3,  increasing  all 
the  dimensions,  except  B  C  and  CD,  by  2  feet,  to  give  room 
to  lay  the  masonry.    (The  reason  for  not  increasing  B  C  and 
CD  will  be  seen  by  drawing  lines  around  the  plan  1  foot 
outside  the  wall  lines.)  Ans.    237.33  cu.  yd. 

(32)  The  length  of  a  building  with  a  plain  roof  is  32  feet, 
and  the  width  is  24  feet.     The  height  of  the  gables  is  f  the 
width.     The  roof  projects  over  the  walls  15   inches  (meas- 
ured on  the  roof)  at  ends  and  eaves,    (a)  How  many  squares 
(100  sq.  ft.)  of  slating  in  the  roof  area  ?    (/;)  The  slates  being 
12  inches  wide  and  exposed  84-  inches  to  the  weather,  how 


many  will  be  required  per  square  ? 


, 


14.12  squares. 
I  (/;)     141+  slates. 

(33)  The    dome  of  a  cupola   is   hemispherical,    and    its 
diameter  is  3  feet.      Deducting  5  square  feet  for  windows, 
etc. ,  what  is  the  area  of  the  remaining  surface  ? 

Ans.   9.14  sq.  ft. 

(34)  Estimating  cast  iron  at  450  pounds  per  cubic  foot, 
what  is  the  weight  per  12-foot  length  of  pipe,  10  inches  inside 
diameter,  the  thickness  being  4-  inch  ?  Ans.    GIG. 5  Ib. 

(35)  The   outside  of  a  circular  tower  28  feet  high   and 
8  feet  in  diameter  is  to  be  shingled,     (a)  How  many  squares 

(100   sq.  ft.)    of   shin- 
gling are  required  ?  (6) 
If   the   shingles  aver- 
age 6  inches  wide  and 
are  laid  5  inches  to  the 
weather,  how  many  will 
be   required  to  the 
square  ? 
A-,   J  («)    "-04  squares. 

s  (  (d)    480. 
(36)    At.  2G  pound  per 
cubic  inch,  what  is  the 
weight  of  the  cast-iron 
FIG  1S  base  shown  in  Fig.  15  ? 


8 


GEOMETRY  AND  MENSURATION. 


Figure  the  4  ribs  as  plain,  with  no  allowance  for  swelled 
parts  nor  deduction  for  holes.  The  corners  of  the  base  are 
rounded  to  a  2-inch  radius.  Ans.  82.32  Ib. 

(37)  A  bridge  pier  28  feet  high  is  12  ft.  X  30  ft.  at  the 
'base,  and  7  ft.  X  22  ft.  at  the  top.     What  will  be  the  cost  of 
the  masonry  at  $12  per  cubic  yard  ?  Ans.   13,115.20. 

(38)  A   cast-iron   ball   which   will   weigh   25   pounds  is 
required.     Figuring  cast  iron  at  .26  pound  per  cubic  inch, 
what  will  be  the  diameter  of  the  ball  ? 

Ans.   5. 68  in.  or  5||  in.,  nearly. 

(39)  Estimating   steel   at 
.283  pound   per  cubic  inch, 
what  will  be  the  weight  per 
foot  of  the  column  shown  in 
Fig.   16,   in  which  (a)  is  an 
enlarged  section  of  a  ' '  chan- 
nel "  ?     Neglect  the   curved 
corners    and    edges  on    the 
channels.  Ans.   62.83  Ib. 

(40)  The   load  on  a  col- 
umn  is  96,000  pounds.     Al- 
lowing a  safe  pressure  of  3 

tons  (of  2,000  Ib.)  per  square  foot  on  the  soil,  and  10  tons 
per  square  foot  on  the  brick  pier,  what  must  be  (a)  the 
dimensions  of  the  square  footing  area,  and  (b]  the  dimen- 
sions of  the  square  column  base  ? 

Ans  \  W     4  ft  S(luare- 
'  (  (b)     2.19  ft.  square. 

(41)  In  a  roof  truss  one  of  the  tie-rods  must  sustain  a 
load  of  27,500  pounds.     The  safe  stress  being  10,000  pounds 
per  square  inch,  what  must  be  the  diameter  of  the  rod  ? 

Ans.   1$  in. ,  nearly. 

(42)  A  square  pyramidal  monument  is  3  feet  square  at 
the  base,  1  foot  square  at  the  top,  and  is  18  feet  high.     It 
is  capped  by  a  pyramid  1  foot  square  and  2  feet  high.  .   If 
the  stone  is  granite,  weighing  170  pounds  per  square  foot, 
what  is  the  total  weight  ?  Ans.   13,373$  Ib. 


FIG.  16. 


ARCHITECTURAL  ENGINEERING. 

(ARTS.   1-134.   SEC.   5.) 


(1)  At  what  point  does  the  greatest  bending-  moment 
occur  in  a  cantilever  beam  ? 

(2)  What  live  load  would  you  use  in  designing  the  floor 
of  a  theater  ? 

(3)  In     selecting    I 
beams,  what  is  consid- 
ered   good    practice   in 
regard  to  the  depth,  so 
as  to  avoid  excessive  de- 
flection ? 

(4)  By  means  of  the 
method  of  the  polygon 
of  forces,  determine  the 
resultant  of  the  several 
forces  shown  in  Fig.  1. 

(5)  Explain  what  is 
meant  by  the  horizontal 

and  vertical  components  of  an  oblique  force. 

(6)  It  is  required  to  span  an  opening  25  feet  wide  in  a 
solid  brick  wall  24  inches  thick,  using  two  I  beams,  side  by 
side.     The  wall  is  laid  in  lime  mortar,  and  the  safe  unit  fiber 
stress   of   the   material   composing  the   I   beams   is   15,000 
pounds;  what  should  be  the  size  of  the  beams? 

12  in.  30.4  lb.,  or 


FIG.  l. 


Ans. 


(  15  in.  41.2  lb. 


2  ARCHITECTURAL  ENGINEERING.  §  5 

(7)  Explain  the  use  of  separators  placed  between  I  beams. 

(8)  The  span  of  a  beam  is  32  feet.     For  three-quarters  of 
this  distance  from  the  left-hand  support  it  is  loaded  with  a 
uniformly  distributed  load  of  6,000  pounds.     At  distances 
of  9  feet  and  14  feet  from  the  right-hand  support  are  located 
concentrated  loads  of  4,500  pounds  and  8,100  pounds,  respect- 
ively.    At  what  distance  from  the  left-hand  end  does  the 
shear  change  sign  ?  Ans.   18  ft. 

(9)  In  what  way  do  the  loads  upon  the  foundations  of  an 
office  building  and  a  storage  warehouse  differ  ? 

(10)  A  square  yellow-pine  column  18  feet  long  must  sup- 
port a  load  of  103,900  pounds;  if  a  factor  of  safety  of  5  is 
used,  what  will  be  the  size  of  this  column  ? 

Ans.   12  in.xl2  in. 

(11)  What  is  the  shear  between  the  points  c  and  b  on  a 
beam  loaded  as  shown  in  Fig.  2  ?  Ans.  800  Ib. 


12-O- 


,     „  c  ~  3000  76. 

h — 10-0—  — H 

-5-0- ^4000  Ib     2000lb] 


-\4000lb 


20'-0- 


FIG.  2. 

(12)  The  length  of  a  beam  is  50  feet,  and  it  overhangs  the 
right-hand  support  10  feet ;  from  the  overhanging  end  there 
is  suspended  a  weight  of  10,000  pounds;  15  feet,  25  feet,  and 
28  feet  from  the  left-hand  support  are  loads  of  9,000,  11,000, 
and  19,000  pounds,  respectively.     What  is  the  right-hand 
reaction?  Ans.  36,050  Ib. 

(13)  Explain  what  is  meant  by  a  reaction. 

(14)  State  the  conditions  demanded  for  good  castings  to 
be  used  in  building  operations. 

(15)  Explain  wherein  the  design  of  the  cast-iron  column 


§  5  ARCHITECTURAL  ENGINEERING.  3 

shown  in  Fig-.  3  is  faulty.      Redesign  the  cap  and  base  to 
meet  the  requirements  of  good  practice. 


(16)  What  is  the  relation  between   the  external  forces 
acting  on  a  beam,  when  the  beam  is  in  equilibrium  ? 

(17)  The    concentrated   loads   upon   a   beam    are   8,000, 
7,000,  and  9,000  pounds.     The  reaction  A',  is  12,000  pounds. 
What  is  the  magnitude  of  the  reaction  A'a  ?     Ans.    12,000  Ib. 

(18)  What  safe  uniformly  distributed  load  will  a  granite 
lintel  20  inches  deep  and  25  inches  wide  sustain,  the  span 
being  5  feet  ?  Ans.  20  T. 

/lOOOlb.pcr  lineal  foot. 


(19)  In  the  trussed  beam  shown  in  Fig.  4,  what  is  (a)  the 
tension  in  .the  camber  rod  ?  (b)  the  compression  on  the 
trussed  beam?  A_s  \  (a)  50,089  Ib. 

(  (b)     55,000  Ib. 


Ans. 


4  ARCHITECTURAL  ENGINEERING.  §  5 

(20)  The   area   of   a   structural  steel  angle  is  6    square 
inches.     What  will  be  the  safe  working  tensile  stress  upon 
this  angle,  providing  a  factor  of  safety  of  3  is  adopted,  and 
the  ultimate  tensile  strength  of  the  material  is  60,000  pounds  ? 

Ans.  120,000  Ib. 

(21)  The  compressive  stress  upon  a  short  oak  block  12 
inches   square   is   135,360   pounds.     What  is  the  factor  of 
safety?  Ans.   3.83. 

(22)  What  factor  of  safety  would  you  use  if  you  were 
designing  a  stone  lintel  to  support  a  given  load  ? 

(23)  When  is  any  structure  in  equilibrium  ? 

(24)  A  girder  of  30-foot  span  is  trussed  at  the  center  by 
camber  rods  and  a  strut.      The  depth  of  truss  from  the 
center  of  the  girder  to  the  center  of  the  rods  is  2  feet ;  if  the 
beam  is  loaded  with  a  uniformly  distributed  load  of  2,500 
pounds  per  lineal  foot,  (a)  what  is  the  stress  on  the  rods  ? 
(b)  What  is  the  compressive  stress  on  the  beam  ?     (c)  What 
is  the  stress  on  the  central  strut  ?  r  (a)     177,337  Ib. 

Ans.  ]   (b)     175,781  Ib. 
(  (c)     46,875  Ib. 

(25)  A  roof  covering  is  composed  of  Spanish  tile  laid 
upon  3-inch  spruce  sheathing.      Between  the  tile  and  sheath- 
ing there  is  placed  2  layers  of  roofing  felt.     What  is   the 
weight  per  square  foot  of  this  covering  ?  Ans.   15  Ib. 

(26)  Explain  the  difference  between  live  and  dead  loads. 

(27)  Calculate  the  square  of  the  radius  of  gyration  for 
the  section  of  a  cylindrical  column  10  inches  outside  diam- 
eter, metal  f  inch  thick.  Ans.   10.77. 

(28)  What  is  the  resisting  moment  of  a  12"  X  1"  wrought- 
iron  plate,  using  a  unit  fiber  stress  of  10,000  pounds  ? 

Ans.  240, 000  in. -Ib. 

(29)  The  span  of  the  15-inch  steel  I  beams  used  in  the 
floor  of  a  fireproof  building  is  23  feet,  and  the  beams  are 
spaced  4  feet  center  to  center.     The  live  load  is  200  pounds 
per  square  foot  of  floor  surface.     The  floor  to  be  supported 


§5  ARCHITECTURAL  ENGINEERING.  5 

by  a  4-inch  brick  segment  arch,  having  a  rise  of  5  inches, 
laid  in  cement  with  the  necessary  concrete  filling,  over  the 
top  of  which  is  laid  a  1-inch  yellow-pine  floor.  The  flooring 
is  nailed  to  2"  x  3 "  sleepers,  embedded  in  the  concrete.  Using 
a  unit  fiber  stress  of  15,000  pounds,  find  the  weight  of  beam 
to  be  used.  In  calculating  the  amount  of  the  dead  load, 
disregard  the  weight  of  the  beams. 

Ans.    15  in.   =  50.9  Ib. 

(30)  The  length  of  a  beam  is  30  feet,  and  its  only  support 
is  at  the  center ;  at  the  left-hand  end  is  a  load  of  (JO  pounds, 
and  9  feet  from  the  left-hand  end  is  a  load  of  80  pounds. 
What  will  be  the  load  required  at  the  right-hand  end  to  pre- 
vent the  beam  from  rotating  around  its  support  ?    Ans.  92  Ib. 

(31)  Explain  what  is  meant  by  (<7)  neutral  axis;  (/;)  resist- 
ing moment. 

(32)  What  would  you  consider  a  safe  live  load  to  be  used 
in  designing  a  country  dwelling  ? 

(33)  Calculate  the  reactions  A,  and  A.^  of  a  beam  loaded 
as  shown  in  Fig.  5.  (  At  20,857}  Ib. 

(  A.,  18,342f  Ib. 


(34)  When  a  beam  is  loaded  with  several  concentrated 
loads,  where  does  its  greatest  bending  moment  occur  ? 

(35)  State  some  of  the  advantages  and  disadvantages  of 
cast-iron  columns. 

(36)  A  cantilever  beam   securely  fastened   into  a  wall 
extends  8  feet  from  the  point  of  support ;  it  is  loaded  with  a 


6 


ARCHITECTURAL  ENGINEERING. 


uniformly  distributed  load  of  500  pounds  per  lineal  foot. 
What  is  the  maximum  bending  moment  in  foot-pounds  ? 

Ans.   16,000  ft. -Ib. 

(37)  The  floor  of  a  factory  building  is  60  ft.  X  290  ft. ; 
what  will  be  the  probable  entire  weight  due  to  the  live  load 
upon  this  floor  area  ?  Ans.  2,610,000  Ib. 

(38)  The  uniformly  distributed  load  upon  a  beam  is  90, 000 
pounds;  the  beam  is  supported  at  both  ends.     What  is  the 
maximum  shear  upon  the  beam?  Ans.  45,000  Ib. 

(39)  What  is  meant  by  the  shear  on  a  beam  ? 

(40)  Explain  why  a  factor  of  safety  is  used  in  designing 
any  structure. 

(41)  A  hollow  cast-iron  column  is  6  inches  in  diameter 
outside,  the  metal  is  £  inch  thick,  and  the  length  of  the 
column  10  feet.     Using  a  factor  of  safety  of  6,  what  safe 
load  will  this  column  support  ?  Ans.  78,000  Ib. 

(42)  The  stress  upon  a  steel  bar  is  80,000  pounds,  and 
the  sectional  area  of  the  bar  is  5  square  inches.    What  is  the 
unit  stress?  Ans.  16,000  Ib. 

(43)  What  safe  load  will  a  brick  pier  3  ft.  x4  ft.  X 10  ft. 
high  support,  providing  the  pier  is  laid  in  lime  mortar  ? 

Ans.   172, 800  Ib. 


1OOO  Ib.lineal  foot. 


FlG.  6. 


(44)  A  beam  is  loaded  as  shown  in  Fig.  6 ;  (a)  what  in 
round  numbers  is  the  greatest  bending  moment,  and  (£)  at 
what  distance  from  the  right-hand  end  does  it  occur  ? 

((«)    105,800  ft. -Ib. 
AnS'((£)    13ft.  4  in. 


ARCHITECTURAL  ENGINEERING. 


(45)  If  a  factor  of  safety  of  5  is  used,  what  must  be  the 
thickness  of  metal  in  a  16-inch  column  (outside  diameter),  24 
feet  long-,  to  carry  a  load  of  421,000  pounds  ?  Ans.  1  in. 

(46)  The  footing  under  a  brick  pier  rests  upon  a  founda- 
tion soil  of  stiff  clay ;  if  the  footing  is  5  feet  square,  what  load 
in  pounds  will  it  safely  carry  ?  Ans.  125,000  Ib. 

(47)  In  a  beam  uniformly  loaded,  and  supported  at  the 
ends,  where  is  the  greatest  bending  moment  ? 


I 


(48)  Draw  a  dead-load  stress  diagram  for  the  truss  shown 
in  Fig-.  7. 

(49)  A  beam  has  a  span  of  30  feet,  and  is  loaded  with  a 
uniformly  distributed  load  of  2,000  pounds  per  lineal  foot; 
what  is  the  greatest  bending  moment  in  inch-pounds  upon 
the  beam  ?  Ans.  2,700,000  in.-lb. 

(50)  What  safe  uniformly  distributed  load  will  a  12"  X  16" 
yellow-pine  beam,  30  feet  long,  carry,  using  one-quarter  of 
the  value  of  the  modulus  of  rupture  for  a  working  stress  ? 

Ans.  20,800  Ib. 

(51)  What  will  be  the  difference  in  weight  between  a 
^-inch  thick  slate  roof,  laid  upon  1-inch  hemlock  sheathing, 
covered  with  two  layers  of  roofing  felt,  and  a  4-ply  slag  roof 
laid  upon  2-inch  spruce  sheathing?  Ans.  If  Ib. 

(52)  A  square  granite  capstone,   on   brickwork   laid   in 
Rosendale .  cement  mortar,  is  required  to  support  a  load  of 
1,000,000  pounds.     Compute  the  dimensions  of  the  stone. 

Ans.  6  ft.  10  in.  square. 


8  ARCHITECTURAL  ENGINEERING.  §  5 

(53)  Design  the  connection  of  an  8-inch  I  beam  with  a 
beam  15  inches  in  depth,   making  the  top  surface  of  the 
two  beams  flush,  and  using  standard  framing. 

(54)  A  12-inch  I  beam  has  an  area  of  14  square  inches; 
what  is  its  section  modulus?  Ans.  52.5. 

(55)  What  will  be  the  allowable  load  on  a  4"  X 12"  spruce 
column  10  feet  long,  if  a  safety  factor  of  4  is  used  ? 

Ans.  25,200  Ib. 

(56)  Calculate  the  full  panel  wind  loads  for  a  100-foot 
span  roof  truss;  the  trusses  are  20  feet  apart  from  center  to 
center,  and  have  a  rise  of  6  inches  per  foot  of  span;  the 
rafter  members  are  supported  at  both  ends,  and  have  three 
intermediate  supports,  placed  so  as  to  divide  them  into  four 
equal  panels.  Ans.  6,624  Ib. 

(57)  Design  a  foundation  pier  to  support  a  column  which 
carries  a  load  of  200,000  pounds;  the  soil  is  a  stiff,  dry  clay, 
and  the  body  of  the  foundation  is  a  good  rubble  masonry  laid 
in  Portland  cement  mortar;  the  capstone  under  the  column 
is  granite. 

(58)  A  tension  member  15  feet  long,  under  an  excessive 
load  is  stretched  T3¥  of  an  inch ;  what  is  the  unit  strain  on 
this  rod  ?  Ans.  .00104  in. 

(59)  What  size  of  steel  I  beam  will  be  required  to  fulfil 

-4000  Ib. 
800  Ib.per  lineal  foot. 


R, 

the  conditions  shown  in  Fig.  8,  if  a  unit  fiber  stress  of  15,000 
pounds  is  used  ?  Ans.  15  in.  41.2  Ib. 

(60)  If  a  structural  steel  bar,  which  has  an  ultimate  ten- 
sile strength  of  60,000  pounds,  is  suspended  from  one  end, 
how  long  must  it  be  to  break  with  its  own  weight  ? 

Ans..  17,668  ft. 


§  5  ARCHITECTURAL  ENGINEERING.  !) 

(61)  Explain  what  is  meant  by  (<?)  unit  stress;   (/>)  unit 
strain. 

(62)  What  are  Newton's  three  laws  of  motion  ? 

(63)  (a)  What   is   the   safe   load    on    a    east-iron    column 
10  inches  square,  outside,  and   20  feet  long,  if  a  factor  of 
safety  of  5  is  used,  the  metal  being  1  inch  thick  ?    (/;)  Design 
the  base  and  also  its  foundation,  using  a  limestone  cap ;  the 
body  of  the  foundation   to  be  composed  of  brick  masonry 
laid  in  Portland  cement  mortar;  the  footing  of  the  founda- 
tion to  be  Portland  cement  concrete,  resting  on  a  good,  dry 
clay,  capable  of  supporting  2J-  tons  per  square  foot. 

Ans.   (a)  268,700  Ib. 

(64)  Draw  the  wind  and  dead  load  stress  diagram  for  the 
truss  shown  in  Fig.  !». 


Fin.  0. 


(65)  Through  what  lever  arm  will  25  pounds  be  required 
to  act,  to  produce  a  moment  of  275  foot-pounds  ? 

Ans.   1 1  ft. 

(66)  In  a  simple  beam,  what  relation  does  the  shear  bear 
to  the  bending  moment  ? 

(67)  What  is  considered  as  the  maximum  safe  deflection 
for  beams  carrying  plaster  ceilings  ? 

(68)  What  factor  of  safety  would  you  deem  advisable  to 
use   in   structures   composed   of    (a)  structural    steel  ?    (/>) 
wrought  iron  ? 

1-34 


10 


ARCHITECTURAL  ENGINEERING. 


(69)  Using   an   ultimate   unit   stress  of    60,000   pounds, 
what  will  be  the  ultimate  tensile  strength  of  a  2"x%"  struc- 
tural-steel bar  ?  Ans.  30,000  Ib. 

(70)  Design  a  wooden  roof  truss,  the  frame  diagram  of 
which  is  shown  in  Fig.  10,  disregarding  the  pressure  due  to 


FIG.  10. 


the  wind.  The  truss  is  composed  of  yellow  pine 
wrought-iron  tension  rods,  and  a  factor  of  safety  of  5 
be  used. 

v 

Explain  the  action  of  wind  upon  roofs. 


with 
is  to 


FIG.  11. 


(72)     Draw  the  vertical-load  stress  diagram  for  the  truss 
shown  in  Fig.  11. 


ARCHITECTURAL  ENGINEERING. 

(ARTS.   1-90.     vSEC.  6.) 


(1)  What  is  the  moment    of   inertia  of    a  circular  area 
4  inches  in  diameter  (<7)  with  respect  to  an  axis  through  its 
center,  and  (/>)  with  respect  to  a  parallel  axis  8  inches  from 
its  center  ? 

(  (a)     12.500. 
Ans.  - 

(  (/>)     81i;.7'.». 

(2)  Calculate  the  moment  of  inertia  of  the  column  section 
shown  in  Fig.    1   (a)  with  respect  to  the  axis  a  />,   and  (/>) 
with     respect    to    the   axis   c  d.  c 

(c)  What    is   the   square    of   the    ^^^^^^^_ '  ^^^^^^^ 
least    radius    of  gyration   of  the 
section  ? 

(  (a)     456.0912. 
Ans.  J  (/;)      123.3042. 

I  (c)      0.20. 

(3)  If  made  of  structural  steel 
with    an    ultimate    compressive 
strength   of    52,000   pounds   per 
square  inch,  and  a  factor  of  safety 
of  5  is  required,  what  safe  load, 
in  round  numbers,  will  a  column 

18  feet  long,  with  hinged  ends  and  with  the  section  shown 
in  Fig.  1,  carry?  Ans.    144,800  Ib. 

(4)  A  15-inch   60-pound  I  beam  24  feet  long  carries  a 
uniformly  distributed  load  of  1,500  pounds  per  lineal  foot. 
What,  in  round  numbers,  is  the  greatest  unit  fiber  stress  ? 

Ans.    15,240  Ib.  per  sq.  in. 


ARCHITECTURAL  ENGINEERING. 


6 


(5)  (a)  What  conditions  should  be  considered  in  choosing  a 
type  of  column  for  a  given  purpose  ?     (b)  Why  is  it  some- 
times better  to  use  a  column  section  in  which  the  distribu- 
tion  of   the   material   is  not  the   most  economical  from  a 
theoretical  point  of  view  ? 

(6)  A  plate  girder  supports  a  floor  surface  24  feet  by  18 
feet,  on  which  the  total  dead  and  live  load  is  400  pounds 
per  square  foot.     If  at  each  end,  the  girder  is  connected  to 
a  column  with   f-inch   rivets   through   ^--inch   connection 
angles,  and  an  allowable   fiber  stress  in  tension  of  12,000 
pounds  per   square   inch    is   used,    how   many  rivets    are 
required  to  support  each  end  of  the  girder  ? 

Ans.  24  rivets. 

(7)  (a]  What  is  the  maximum  allowable  pitch  of  rivets  in 
compression  members  ?     (b)  What  is  the  minimum  distance 
that  a  f-inch  rivet  may  be  placed  from  the  end  of  a  ^--inch 
plate  ?  Ans.   (b)     1^-  in. 

(8)  (a)  What  is  understood  by  the  term  camber  as  applied 
to  a  roof  truss  ?     (b)  What  is  the  effect  of  camber  on  the 
strength  of  the  members  of  a  truss  ? 

(9)  A  steel  rod  7£  inches  long  and  ^  inch  in  diameter  is 
elongated  .009  inch  by  a  pull  of  7,000  pounds;    what,  in 
round  numbers,  is  the  modulus  of  elasticity  ? 

Ans.  29,709,000. 

(10)  (a)  Calculate  the  moment  of  iner- 
tia of  the  section  shown  in  Fig.  2  with 
respect  to  an  axis  parallel  to  the  upper 
edge  and  passing  through  the  center  of 
gravity  of  the  figure,     (b)  What  is  the 
least  radius  of  gyration  of  the  section  ? 

(a)  280.469. 

(b)  1.55. 

(11)  (a)  What  advantage    does    the 
box  section  plate  girder  have  over  the 

other  forms?       (b)  What    serious    disadvantage     does     it 
have  ? 


Ans. 


§  6  ARCHITECTURAL  ENGINEERING.  3 

(12)  (a)  How  is  the  distribution  of  the  stresses  in  a  plate 
girder  assumed  to  differ  from  those  in  a  beam  composed  of 
a  single  piece  ?     (//)  What  part  of  the  girder  is  assumed  to 
resist  the  shearing  stresses  ?     (c]  What  part  of  the  girder  is 
assumed  to  resist  the  stresses  due  to  the  bending  moment  ? 

(13)  Calculate  the  thickness  of  the  web-plate  for  the  plate 
girder  in  example  6,  if  the  depth  of  the  girder  is  3o  inches. 
Each  pair  of  the  end  stiffeners  is  fastened  to  the  web-plate 
by  nine  f-inch  rivets,  and  the  allowable  resistance  of  the 
material  to  shearing  is  12,000  pounds  per  square  inch. 

Ans.  -|  in. 

(14)  A  uniformly  loaded  plate  girder  of  75-foot  span  has 
a  flange  with  two  6"x4"x4-"  angles  and  four  8"x|"  cover- 
plates;  there  is  a  row  of  f-inch  rivets  joining  these  plates  to 
each  flange  angle,  and  a  single  row  of  f-inch  rivets  joining 
the   flange  angles    to    the   web-plate,    the    rivets   being   so 
arranged  that  the  section  to  be  deducted  is  that  of  two  rivet 
holes  for  each  of  the  flange  plates  and  flange  angles ;  begin- 
ning with  the  outside  plate,  what  are  the  theoretical  lengths 
of  the  plates?  f  1st  plate,  27  ft.  0  in. 

2d    plate,  30  ft.  3  in. 
"*  3d    plate,  48  ft.  1  in. 
4th  plate,  55  ft.  G  in. 

(15)  (a)  What  is  the  elastic  limit  ?     (/>)  When  is  a  body 
said  to  have  a  permanent  set  ?     (c)  Are  the  materials  used 
in  building  construction  thought  to  be  perfectly  elastic  ? 

(16)  A  10-inch  25-pound  structural-steel  I  beam  with  a 
span  of  14  feet  supports  a  floor  surface  24  inches  wide,  on 
which  the  total  dead  and  live  load  is  150  pounds  per  square 
foot;  what  is  the  deflection  of  the  beam  ?  Ans.  .073  in. 

(17)  The  yellow-pine  rafter  member  of  a  composite  pin- 
connected  roof  truss  is  52  feet  long.      It  is  divided  into  four 
equal  panels,  and  the  roof  and  wind  exert  a  pressure  equal 
to  a  uniformly  distributed  load  of  GOO  pounds  per  lineal  foot 
of  the  rafter.     The  stress  diagram  shows  a  maximum  com- 
pressive  stress  on  the  rafter  of  56,000  pounds;  if  the  depth 


4  ARCHITECTURAL  ENGINEERING.  §  6 

of  the  rafter  is  10  inches  and  a  factor  of  safety  of  4  is  used, 
what  must  be  the  thickness  ?  Ans.  10  in. 

(18)  A    white-pine    beam   of    rectangular    cross-section 
carries  a  uniformly  distributed  load  of  50  pounds  per  lineal 
foot.     If  the  beam  is  8  inches  by  14  inches  and  16  feet  long, 
how  much  more  will  it  deflect  when  the  short  side  is  vertical 
than  when  the  long  side  is  vertical  ?  Ans.   .0831  in. 

(19)  (a)  What  type  of  roof  truss  is  most  commonly  used 
for   buildings    of   moderate   span  ?      (b)  What   difficulty   is 
encountered  in  constructing  the  stress  diagram  for  this  truss  ? 
(c)  Explain  the  method  by  which  the  diagram  is  drawn  so  as 
to  find  the  stresses  at  the  joint  where  the  difficulty  occurs. 

(20)  Why  are  some  of  the  members  of  a  structural-steel 
roof  truss  made  heavier  than  is  demanded  by  the  stresses 
they  must  withstand  ? 

(21)  (a)  What    is  a   flitch-plate    girder?     (b}  What   are 
some  of  its  advantages  in  comparison  with  a  simple  steel 
beam  ?     (c)  In    the    design  of    a  flitch-plate  girder,    what 
should  be  the  relation  between  the  different  members  ? 


What  is^the  reason  for  using  flange  plates  of  differ- 
ent lengths  in  the  construction  of  a  plate  girder  ? 

(23)  Name  the  different  methods  that  may  be  used  in 
calculating  the  pitch  of  the    rivets   connecting  the    flange 
angles  with  the  web-plate  of  a  plate  girder. 

(24)  By  means  of  a  diagram,  determine  the  lengths  of 


* 

!•* 

9 
c 

a 

6                         1 

o 

c 

c                          s 

i 

d                   «c 

e 

2 

at 

f                                               > 

»1 

r                                   i 

-/*- 

<•          /£           >• 

<  /s'  *• 

<  /&'  —  - 

/ 

<  /4  >- 

+-*± 

FIG.  3. 


the  flange  plates  for  a  plate  girder   loaded    as    shown    in 
Fig.  3.     The  flange  is  made  up  of  three  10"  X  ¥  plates  and 


ARCHITECTURAL  ENGINEERING. 


ttotesfor^ffiveis. 


FIG.  4. 


two  4"x4"xf"  angles,  connected 
as  shown  in  Fig.  4. 

f  Outside  plate,  30  ft.  0  in. 
Ans.  -j  Middle  plate,   40  ft.  1)  in. 

(  Inner  plate,  03  ft.  0  in. 
(25)  The  depth  of  the  girder 
in  the  last  example  is  5  feet;  (<?) 
calculate  the  size  of  the  angles 
required  for  the  end  stiffeners,  if 
two  angles  are  used  and  a  com- 
pressive  fiber  stress  of  12,000  pounds  per  square  inch  is 
allowed ;  (b}  if  the  section  cut  out  for  fifteen  holes  for  |-inch 
rivets  is  deducted  from  its  total  depth,  calculate  the  thick- 
ness of  the  web-plate,  allowing  a  shearing  fiber  stress  of 
12,000  pounds  per  square  inch. 

\  (a]     2f "  X  2|"  X  f "  angles  may  be  used. 
(  (/;)     Calculated  thickness,  -^  in. 

(20)  What  is  the  net  section  of  a  0"xO"xTV  angle  from 
which  is  to  be  deducted  the  section  cut  out  for  two  |-inch 
rivets?  Ans.  4.185  sq.  in. 

(27)  (a)  What  is  considered  the  ratio  of  depth  to  span  of 
a  plate  girder  that  should  be  allowed  in  the  best  practice  ? 
(b}  If  a  smaller  ratio  is  desired  in  order  to  meet  conditions 
demanded  by  the  construction  of  a  building,  what  precau- 
tion is  required  in  the  design  ? 

(28)  (a)  What  is  the  usual  practice   in  regard  to  rivet 
spacing  at  the  joints  and  foot  of  a  built-up  column  ?    (b}  Give 

a  practical  rule  for  the  diameter  of 
rivets  in  built-up  columns. 

(29)  What  are  some  of  the  disad- 
vantages of  the  Phoenix  column  sec- 
tion in  building  construction  ? 

(30)  With  a  factor  of  safety  of  4, 
what,  in  round  numbers,  is  the  great- 
est allowable   load    on   a   structural  - 

FIG.  5.  steel  column  24  feet  long  with  fixed 

ends  and  with  the  section  shown  in  Fig.  5  ?      Ans.  1 18,500  Ib. 


6 


ARCHITECTURAL  ENGINEERING. 


6 


(31)  (a)  What  rolled  sections  are  most  often  used  in  the 
construction   of   structural-steel   roof   trusses   of  moderate 
span  ?     (&)  How  are  these  sections  connected  ? 

(32)  (a)  What  is  gained  by  upsetting-  the  ends  of  a  ten- 
sion member  ?     (b)  Would  it  be  economical  to  upset   the 
ends  of  very  short  tension  members  ? 

(33)  What  should  be  the  relation  between  the  size  of  the 
pins  used  in  the  eyes  of  tension  members  and  the  dimen- 
sions of  the  bars  ? 

(34)  Why  is  a  lower  factor  of  safety  allowable  in  the 
design  of  a  roof  than  in  the  design  of  a  bridge  ? 

(35)  Find  (a)  the  horizontal,  (b)  the  vertical,  and  (c)  the 
maximum  bending  moments  on  the  pin  shown  in  Fig.  6. 


L^ 


FIG.  6. 

(d}  If  the  allowable  fiber  stress  is  20,000  pounds  per  square 
inch,  what  must  be  the  diameter  of  the  pin  ? 

(a)      59,250  in.-lb. 

(b}      24, 000  in.-lb. 

(c)      63,926  in.-lb. 


Ans. 


6  ARCHITECTURAL  ENGINEERING. 


(36)  Fig-.  7  shows  the  members  meeting  at  a  joint  of  the 
lower  chord  of  a  structural-steel  roof  truss,  with  the  stress 
in  each  member.  Assuming  an  allowable  stress  per  square 
inch  in  tension  of  12,000  pounds  per  square  inch  and  a  thick  - 


-60OOO  Ib. 


400OO  lb- 


FIG. 


ness  of  gusset  plate  of  T7¥  inch,  (a)  calculate  the  number  of 
f-inch  rivets  required  in  the  end  of  each  member;  (/>) 
make  a  sketch  showing  the  gusset  plate  and  the  arrangement 
of  the  rivets.  A  splice  plate  may  be  used  to  connect  the 
chord  angles  if  required.  I  Member  a,  9  rivets. 

Member  b,  3  rivets. 
Ans.    (a)  \  ., 

v  '    !  Member  c,  3  rivets. 

Member  d,  6  rivets. 

(37)  Calculate  the  moment  of  inertia,  with  respect  to  an 
axis  18  inches  from  its  center,  of  the  section  of  a  hollow 
cylinder   whose   outside    diameter   is    12  inches  and  inside 
diameter  10  inches.  Ans.  11,724.405. 

(38)  Two  o^xS^Xf"  angles,   placed  back  to  back  with 
the  long  legs  parallel  and  1  inch  apart,  are  to  be  used  as  a 
column  with  fixed  ends;  the  length  of  the  column  is  22  feet. 
In  round  numbers,  what  load  will  the  column  carry  with  a 
factor  of  safety  of  4  ?  Ans.   72,000  Ib. 

(39)  The   reaction   at  the  support  of  a  plate  girder  is 
156,000  pounds;  the  depth  of  the  girder  is  48  inches,  and 
there  are  14  holes  for  f -inch  rivets  to  be  deducted  from  the 


8 


ARCHITECTURAL  ENGINEERING. 


(i 


width  of  the  web-plate.  If  an  allowable  shearing  stress  of 
13,000  pounds  per  square  inch  is  used,  what  must  be  the 
thickness  of  the  web  ?  Ans.  .336  in.,  say  -f  in. 

(40)  A  plate  girder,  with  a  span  of  84  feet,  carries  a 
uniformly  distributed  load  of  3,000  pounds  per  lineal  foot; 
the  web-plate  is  T5¥  inch  thick,  and  it  is  divided  into  14  equal 
panels ;  if  f -inch  rivets,  spaced  according  to  the  direct 
vertical  shear,  are  used,  and  the  allowable  stress  in  tension 
is  15,000  pounds  per  square  inch,  what  should  be  the  spa- 
cing, for  the  successive  panels  from  the  support  to  the 
center,  of  the  rivets  connecting  the  flange  angles  to  the 


web  ? 


(41) 
beams 


Ans. 


1st  panel,  3.43  in. 

2d  panel,  4  in. 

3d  panel,  4.8  in. 

4th,  5th,  6th,  and  7th  panels,  6  in. 

A  floor  is  to  be  supported  by  12-inch  31^-pound  I 
with  a  span  of  20  feet  spaced  24  inches  between 


centers ;  the  total  dead  and  live  load  on  the  floor  is  to  be 
480  pounds  per  square,  foot.  What  will  be  the  greatest 
deflection?  Ans.  .54  in. 

(42)  By  means  of  the  principle  of  moments,  calculate  the 
distance  d  of  the  neutral  axis  a  b,  of  the  sec- 
tion in  Fig.  8,  from  the  outer  edge  of  the 
plate.  Ans.  2.137  in. 

(43)  What  is  the  moment  of  inertia  of  the 
section  shown  in  Fig.  8  with  respect  to  the 
axis  a  b  ?  Ans.   64. 567. 

(44)  What  is  the  section   modulus  with 
respect  to  an  axis  perpendicular  to  the  web 
(a)  of  a  10-inch  33-pound  I  beam;  (b)  of  a 
12-inch  20-pound  channel  ?  A        ((a)     32.26. 

L1(*)     20.78. 

(45)  With  a  maximum  fiber  stress  of  12,000  pounds  per 
square  inch,  what  is  the  resisting  moment  of  each  of  the  sec- 
tions in  the  preceding  example  ?  .  j  (a)  387,120  in.-lb. 

"  (  (b)     249, 360  in.-lb. 


§  6  ARCHITECTURAL  ENGINEERING.  9 

(4G)  (a)  Why  should  a  greater  factor  of  safety  be  used  for 
long  columns  than  for  short  ones  ?  (/>)  Give  a  formula  for 
the  factor  of  safety  to  be  used  for  any  column  with  round  or 
hinged  ends. 

(47)  Describe   Osborn's  code  of  conventional  signs  for 
rivets. 

(48)  (a)   In  what  ways  may  a  riveted  joint  fail  ?     (b)  What 
relations  are  assumed  between  the  tensile,  compressive,  and 
shearing  strengths  of  the  metal  in  computing  the  strength 
of  rivets  and  riveted  joints  ? 

(49)  What  kinds  of  stresses  are  most  likely  to  cause  fail- 
ure in  pins  ? 

(50)  (a)  What  rule,  in  respect  to  the  web-plate,  is  some- 
times used  in  calculating  the  flange  area  of  plate  girders  ? 
(b)  What  precautions  should  be  observed   in   splicing  the 
web-plates  of  plate  girders  ? 


INDEX. 


NOTE. — All  items  in  this  index  refer  first  to  the  section  (see  Preface,  Vol.  I)  and  then 
to  the  page  of  the  section.  Thus,  "Burl  925"  means  that  burl  will  be  found  on  page  •£> 
of  section  9. 


A. 

Page. 

A'<r. 

Pti.tr, 

Abstract  numbers  

1 

1 

Arch,  Span  of  

4 

M 

Accidental  load  

5 

38 

Architectural   engineering,    Na- 

Acute angle  

4 

3 

ture  of  

5 

1 

Addition  

1 

4 

Area  of  anv  plane  figure  

4 

3'.) 

"        of  decimals  

1 

88 

"       "  circle  

4 

34 

"         of  denominate  numbers 

2 

15 

"       "  circular  ring  

4 

35 

"        of  fractions  

1 

27 

"       "  ellipse  

4 

as 

"        Rule  for  

1 

8 

"       "  irregular  figure  

4 

40 

"        Sign  of  •  

1 

4 

"       "  parallelogram  

4 

29 

"        table  

1 

5 

"       "  polygon  

4 

30 

Adjacent  angles  

4 

3 

"       "  sector  of  circle  

4 

86 

Aggregation,  Symbols  of  

1 

49 

"       "  segment  of  circle  

4 

37 

Altitude    of    parallelogram     or 

"       "  surface  of  frustum  

4 

49 

trapezoid  

4 

29 

"       "         "         "  pyramid       or 

"       of  prism  or  cylinder  

4 

45 

cone  

4 

47 

"        "  pyramid  or  cone  

4 

47 

"         "  triangle  

4 

8 

inder  

4 

45 

Amount  (Percentage)  

2 

2 

4 

53 

Angle,  Acute  

4 

3 

"       "  trapezoid  

4 

29 

"       Definition  of  

4 

2 

"       "  triangle  

4 

27 

"       Inscribed  

4 

18 

Arithmetic,  Definition  of  

1 

1 

"       Obtuse  

4 

3 

"            Fundamental    proc- 

"       Vertex  of  

4 

2 

esses  of  

1 

4 

Angles  

5 

80 

Avoirdupois  weight  

2 

9 

"       Areas  of  :  Table  

6 

137 

Axis,  Neutral  

0 

1 

"       Connection  

5 

97 

"             " 

5 

70 

"       or  arcs.  Measurement  of 

8 

11 

"      of  svmmetrv  

4 

54 

"       Measurement  of,  by  arcs 

4 

18 

"       Opposite  

4 

3 

B. 

Sec. 

Pagt 

"       Properties  of:  Table  

6 

138 

Base  of  triangle  

4 

8 

"                "            "         "      

6 

140 

"     (Percentage)  

2 

2 

"       Right  

4 

3 

Beam,  Bending  moments  in  

5 

78 

Antecedent  of  a  ratio  

9 

43 

"       Continuous  

5 

02 

Arabic  notation  

1 

2 

"       Definition  of  

5 

02 

Arc  of  circle  

4 

16 

"       Deflection  of  

0 

102 

"    "       "     Length  of  

4 

31 

"       Fixed  

5 

02 

XI 


Xll 


INDEX. 


Sec. 

Page. 

Sec. 

Page. 

Beam  girders  

5 

96 

Cast-iron  columns,  Inspection  of 

K 

60 

"      Relation    between    depth 

Cause  and  effect,  Principle  of... 

9 

54 

of,  and  span  

5 

94 

Center  of  circle  

4 

16 

"      Simple  

5 

62 

"        "  gravitv  

8 

26 

"      Span  of  

5 

62 

"        "         "      of  plane  figures 

6 

26 

"      Stone  

5 

99 

"        "moments  

5 

18 

"      Strength  of  

5 

84 

Chain,  Engineers'  

2 

9 

Beams,  Stresses  in  

5 

69 

"        Gunter's  

* 

8 

"        Trussed  

5 

101 

Channel,    Approximate    section 

"       with  one  strut,  Stresses 

modulus  of  

6 

87 

in  

5 

102 

"         column  

6 

31 

"        with  two  struts,  Stresses 

Channels  

6 

86 

in  

5 

103 

"         Elements  of  :  Table... 

5 

90 

Bearing  value  of  rivets  

6 

45 

"         Properties  of  :  Table.. 

• 

144 

"            "       "        "    

6 

136 

Chord  of  circle  

4 

16 

"             "       "        "      Table  of 

6 

49 

Cipher  

1 

2 

"             "       "        "           "        " 

6 

136 

Circle,  Arc  of  

4 

16 

Bending  moment  

5 

74 

"        Area  of  

4 

34 

"                "       and       moment 

"        Center  of  

4 

16 

of  resistance, 

"        Chord  of  

4 

16 

Relation    be- 

"      Circumference  of  

4 

16 

tween  

6 

14 

"        Definition  of  

4 

16 

Bending  moments,  General  for- 

"       Diameter  of  

4 

16 

mulas  for.. 

5 

81 

"        Radius  of  

4 

16 

"           in  shear,  Re- 

"       Relation     between      cir- 

lation   be- 

cumference and  diame- 

tween   

5 

79 

ter  of  

4 

31 

in    simple 

"        Sector  of  

4 

17 

beams  

5 

78 

"        Segment  of  

4 

16 

Bending  stress  

5 

40 

Circles,  Concentric  

4 

20 

Bending  stresses  

5 

74 

"        Equality  of  

4 

17 

Box  column  *  

6 

31 

Circular  measure,  Explanation  of 

4 

17 

Brace  

1 

49 

"       ring,  Area  of  

4 

35 

"     .. 

3 

2 

Circumference  and  diameter  of 

Bracing  of  roof  trusses,  Lateral 

6 

131 

circle.  Relation 

Brackets  

1 

49 

between  

4 

31 

" 

3 

2 

"            of  circle  

4 

16 

Broken  line  

4 

1 

Circumscribed  polygon  

4 

21 

Buckling  of  web-plate  

6 

63 

Column,  Box  

6 

31 

Building  materials,  Elasticity  of 

6 

102 

"         Channel  

6 

31 

"                  "        Deterioration 

Grey  

6 

32 

of  

5 

47 

"         Keystone  octagonal.  ... 

6 

32 

"                   "         Weight  of.... 

5 

27 

"         Larimer  

6 

31 

"         Latticed  angle  

6 

32 

"         sections  used  in  building 

C. 

Sec. 

Page. 

construction  

6 

30 

Camber  in  roof  truss  

6 

110 

"         splices  and  cohnections 

6 

34 

Cancelation  

1 

19 

"         used  by  Pa.  R.  R  

6 

32 

"            Rule  for  

1 

21 

"         Z-bar  

6 

31 

Cantilever  

5 

62 

Columns  

5 

53 

Capacity,  Measures  of  

2 

10 

Columns  and   connections,    De- 

Cast-iron columns  

5 

56 

tails  of  

6 

33 

"        "           "        Danger    to, 

"         Cast-iron  

5 

56 

from  fire  

5 

61 

"         Conditions    which  af- 

"       "           "       Design  of  

5 

59 

fect  the  choice  of  

6 

26 

"        "           "       Formula  for... 

5 

57 

"         Factors  of  safety  for..  . 

6 

25 

INDEX. 


Xlll 


Sec. 

Page. 

Sec. 

Page. 

Columns,  Formulas  for  

0 

18 

Cubic  measure  

2 

9 

"         Long  

5 

54 

Cubical  contents  of  solid  

4 

44 

"         Method  of  securing  the 

Curved  line  

4 

1 

ends  of  

5 

55 

Cvlinder  

4 

45 

"          Rivet  spacing  in  

6 

:« 

"         Area  of  surface  of  

4 

45 

"         Round-ended  

6 

17 

"         Circular  

4 

45 

"          Short  

5 

58 

Right  

4 

45 

"          Types  of  

6 

16 

l;         Volume  of  

4 

40 

"          with  eccentric  loads... 

6 

22 

*'              "    fixed  ends  

6 

17 

I). 

Sec. 

Page. 

"             "        "          "    

6 

21 

Dates,  To  find  interval  of  time 

"              "    flat  ends  

6 

17 

between  

2 

18 

"              "      "        "     

6 

20 

Dead  load  

5 

27 

"             "    hinged  ends  

0 

17 

Decagon  

4 

0 

6 

18 

Decimal  point  

1 

86 

\Vood  

5 

54 

"         To     express     approxi- 

Common denominator  

1 

26 

mately  as  a  fraction 

Components  of  a  force  

5 

15 

having   a  given   de- 

Composite   pin-connected    roof 

nominator  

1 

48 

truss,  Design  of  

6 

118 

"           "  reduce     a     fraction 

Composition  of  forces  

5 

6 

to  a  

1 

4(3 

Compound  denominate  number 

•2 

8 

"           "  reduce,  to  a  fraction 

1 

47 

proportion  

•2 

55 

Decimals  

1 

3T> 

"                 "             Rule  for 

o 

50 

"         Addition  of  

1 

38 

Compression    members  of    roof 

"         Division  of  

1 

42 

truss  

6 

132 

"         how  read  

1 

30     • 

Compressive  stress  

5 

39 

"         Multiplication  of  

1 

40 

Concentric  circles  

4 

20 

"         Reduction  of  

1 

40 

Concrete  numbers  

1 

1 

"         Subtraction  of  

1 

39 

Conditions  of  equilibrium  

6 

17 

Deflection  of  beams  

0 

102 

Cone  

4 

47 

"          "        "       

0 

104 

"     Altitude  of  

4 

47 

"          "  floor  beams  

5 

110 

"     Frustum  of  

4 

49 

Denominate    number,    Com- 

"    Right  

4 

47 

pound  

2 

8 

"     Slant  height  of  

4 

47 

"             numbers  

2 

7 

"      Surface  of  

4 

47 

"                     "        Addition  of 

o 

15 

"     Volume  of  

4 

48 

"                     "        Division  of 

0 

20 

Consequent  of  a  ratio  

2 

43 

"                     "        Multiplica- 

Contact, Point  of  

4 

19 

tion  of..  . 

2 

19 

Continuous  beam  

5 

02 

"                     "        Reduction 

Connection  angles  

5 

97 

of  

0 

12 

Conventional  signs  for  rivets  

6 

41 

"                     "        Simple  

2 

7 

"                 "       "      rolled 

"                    "        Subtrac- 

shapes 

6 

42 

tion  of... 

2 

17 

Conversion  tables  

4 

25 

"                     "        To  reduce, 

Couplet  of  a  proportion  

2 

47 

to  higher 

"        (Ratio)  

2 

43 

denomi- 

Cross-section of  a  solid  

4 

51 

nations.. 

.7 

13 

Cube  

4 

44 

"                     "        To  reduce, 

"      of  a  number  

2 

23 

to    lower 

"      root  

2 

25 

denomi- 

"        " 

2 

32 

nations.. 

2 

12 

"    Proof  of  

2 

38 

Denominator  

1 

22 

'•'         "    Rule  for  

2 

38 

Depth  of  beam  and  span.  Rela- 

Cubic foot,  Definition  of  

4 

45 

tion  between  

5 

94 

"      inch,  Definition  of  

4 

45 

Design  for  a  large  building  

5 

142 

XIV 


INDEX. 


Design  of  cast-iron  columns  .....      5 

"      "  composite    pin-con- 

nected roof  truss  ----      6 

"      "  foundations  ............      5' 

"      "  roof   truss,   General 
notes  regarding  ...... 

"      "  structural-steel    roof 

truss  ..................      6 

Designing  the  members  of  a 
truss  ..........................      5 

Details    of    structural    columns 
and  connections  ..............      6 

Deterioration  of  material  ........      5 

Diagonal  of  quadrilateral  ........      4 

Diagram  for  simple  frames  ......      5 

"  "    small  roof  truss 

"         Frame 

"         how  lettered  ........  ... 

"         Stress 
Diameter  and  circumference  of 
circle,    Relation     be- 
tween .................       4 

"        of  circle  ................      4 

"         "circle   determined 

from  area  .........      4 

"         "  sphere    determined 

from  volume  ......      4 

Difference  ........................       1 

"         (Percentage)  ..........      2 

Digits  .............................       1 

Direct  proportion  ................      2 


ratio  ...........  4  .......... 

Dividend  .......................... 

Division  ........................... 

"        of  decimals  .............. 

"         "  denominate  numbers 

"         "  fractions  .............. 

"        Rule  for  .................. 

"        Sign  of  ...................      1 

Divisor  ............................       1 

Dodecagon  ........................      4 

Dollars  and  cents,  how  written 

decimally  ..............       1 

"        Sign  for  ..................       1 

Drawings,  how  dimensioned  ____      4 

Dry  measure  ......................      2 

Dynamics  .........................      5 


Page. 
59 

118 
50 

6  131 
124 
139 

33 

47 
29 
116 

5  121 
5  112 
5  113 
5  112 


85 


2 
2 
47 
48 
16 
16 
42 
20 
82 
18 
16 
16 
6 


E.  Sec.  Page. 

Eccentric  loads  on  columns  ......  6       22 

Elastic  limit,  Definition  of  .......  6      102 

Elasticity,  Definition  of  ..........  6      102 

"             Modulus  of  ............  6      Ml 

of  building  materials  6      102 
Elementary  area,  Definition 

of...  68 


Elements  of  usual  sections: 
Table 

Ellipse 

"  Area  of 

"  Perimeter  of 

Elongation,  Ultimate 

Engineers'  chain 

Equality,  Sign  of 

Equiangular  polygon 

Equilateral  polygon 

triangle 

Equilibrium 


Sec.  Page. 

6      135 
4       37 


of  moments. 

"  Stable 

"  Unstable 

Evolution 

Exponent 

Extension,  Measures  of.. 
Extremes  (Proportion). . . 


F.  Sec. 

Factor  of  ignorance 5 

"      "    safety 5 


"     Prime 1 

Factors 1 

"         of  safety  for  columns. ..  6 

"          "        "        "    roof  truss  6 

"          "        "      Table  of 5 

Fathom 2 

Fatigue  of  metals 5 

Figure,  Plain 4 

Figures 1 

"        Local  or  relative  values 

of 1 

"        Similar 4 

"        Simple  value  of 1 

"        Symmetrical 4 

Filler  for  plate  girder 6 

Fink    truss,    Determination    of 

stresses  in 6 

Fixed  beam 5 

Flange  angle 6 

"       plate 6 

"  plates  for  girder  with  con- 
centrated loads  in 
uniformly  distrib- 
uted load 6  85 

"  "    Graphical     method 

of   determining 

length  of 6       74 

"  "    Length  of 6       72 

"    Rivet  spacing  in 6       96 

"       stresses 6       08 

Flanges  of  girder,  how  propor- 
tioned ...  6       69 


38 

42 

9 

4 

7 
6 


16 
19 
16 
16 
25 
23 
8 
47 

Page. 
4C 
46 
20 
19 
25 
132 
47 
11 
47 

6 

2 

2 
55 

2 
54 

57 

110 
62 
57 
57 


Flanges  of  girder  with  concen- 
trated     loads, 
Graphical     deter- 
mination of  length 
of  

Sec. 

6 
6 
6 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
3 
4 
3 
6 

5 
5 
5 
1 
1 
1 
1 
1 
1 
1 
1 

1 

1 

1 
1 
1 
1 
1 
1 
8 
1 

1 
5 

INDEX. 

Page. 

Frame  diagram  

Sec. 
5 
5 
4 

4 

4 
5 

Sec. 
2 
2 
o 
2 
4 
6 

6 
6 
6 
6 
6 
5 
6 
6 

6 

6 

6 
5 

5 
5 

5 
5 
5 
5 

6 

5 
5 
5 
6 

XV 

Page. 
112 
98 
49 

40 

50 
23 

Page. 
6 
11 
11 
11 
1 
00 

97 
67 
106 
57 
106 
96 
90 
58 

78 

78 

74 

122 

116 
121 

133 
126 
123 
105 

5 

11 
111 
26 
32 

78 
67 
106 
110 
18 
18 
3 
3 
15 
6 
6 
16 
6 
11 
14 
9 
1 
51 
10 
18 

48 
48 
50 
22 
23 
25 
23 
23 
33 
47 
46 

23 

24 

24 
23 

27 
32 
31 
23 
40 
29 

27 
113 

Framing  for  I-beam  connections 
Frustum  of  pyramid  or  cone  

of  sur- 
face of 

"         "  plate  girder  

Floorbeams,  Deflection  of  

ume  of 
Fulcrum  

Force,  Components  of  

G. 

Gain  or  loss,  per  cent      

"        Effect  of. 

"        Lever  arm  of  

Gallon,  Cubic  inches  in  

"        Representation  of  

"         Weight  of  

Forces,  Composition  of 

Gallons  in  cubic  foot  

Geometry,  Definition  of  

Girder,  Depth  of  

"         Polygon  of.       .         

"         design,    Practical  prob- 
lems in  

"         Resolution  of 

"         Flanges  of  

Formula,  Definition  of. 

"         Flitch-plate  

"           Prismoidal  

"         Plate  

"           Rankine-Gordon 

Formulas  for  steel  columns  
Foundation  materials,  Strength 
of  

Girders,  Beam  

"         Rivet  spacing  in  

"         Usual  sections  of  

Foundations  

"         with    concentrated 
loads,  Application  of 
graphical  method  to.  . 
Graphical  determination  of  flange 
plates  of  girder  with 
concentrated  loads.  .  . 
Graphical  determination  of 
length  of  flange  plates 
"           diagram  for  jib  crane 
"           "simple 
frames.  .  . 
"                "          "   small  roof 
truss  .... 
"                 "           "wooden 
truss,  80- 
foot  span 
"                "         of  roof  truss, 
for  church.  . 
"                "         of  roof  truss, 
40-foot  span 
"         method  of  computing 
stresses.. 
"                "         "     locating 
neutral 
axis  

"             Design  of  

Fraction  

"         Improper  

"         Lowest  terms  of  

"        Proper  

"         Terms  of  

"         To  invert  a  

"         To  reduce  a  decimal  to 
"          "         "        to  a  decimal 
"          "         "        to    a    higher 
term  

"          "         "       to   an   equal 
fraction 
with        a 
given    de- 
nominator 
"          "         "        to       lower 
terms  
"        Value  of  

Fractions,  Addition  of  

u            Division  of  

"            Multiplication  of  
**            Reduction  of... 

"            Roots  of  

"            Subtraction  of  

"         methods  of  computing 
moments  of  inertia.  . 
"         statics                    

"            To  reduce,  tocommon 
denominator  

Frame  and  stress  diagrams,  how 
lettered  

Gravity,  Center  of  

1-35 

XVI 


INDEX. 


Gunter's  chain 2 

Gusset  plate 6 

Gyration,  Radius  of 5 

"        " 6 

H.  Sec. 

Heptagon 4 

Hexagon 4 

Horizontal  line 4 

Hypotenuse 4 

I.  Sec. 
I    beam,    Approximate    section 

modulus  of 5 

"       "        Connections,  Standard 

framing  for 5 

I  beams 5 

"      "      Elements  of :  Table 5 

"      "      Properties  of :  Table....  6 

"      "       Separators  for 5 

Improper  fractions,  To  reduce,  to 

mixed  numbers 1 

Inches,  To   reduce,   to   decimal 

parts  of  a  foot 1 

Index  of  a  root 2 

Inertia,  Moment  of 6 

Inscribed  angle 4 

"          polygon 4 

Inspection  of  cast-iron  columns..  5 

Integer 1 

Intensity  of  stress 5 

Intersection,  Point  of '. .  4 

Inverse  proportion^ ". 2 

"           2 

"       ratio 2 

Involution , 2 

Isosceles  triangle 4 


Page. 

'Sec. 

Page. 

8 

Limit,  Elastic  

6 

102 

128 

Line  

4 

1 

57 

"    Broken  

4 

1 

15 

"    Curved  

4 

1 

"    Horizontal  

4 

2 

Page. 

"    Right  

4 

1 

6 

"    Straight  

4 

1 

6 

"    Vertical  

4 

2 

2 

Linear  measure  

» 

8 

8 

"         Surveyors'  

g 

8 

Lines,  Parallel  

4 

1 

Page. 

Liquid  measures  

8 

10 

Live  load  

6 

31 

87 

Load,  Accidental  '.  

5 

38 

"      Dead  

6 

27 

98 

"      Live  

5 

31 

86 

Loads  carried  by  structures  

5 

27' 

89 

"       Moments  due  to  

5 

78 

143 

"       Snow  and  wind  

5 

33 

94 

Local  and  relative  values  of  fig- 

ures   

1. 

2 

25 

Long  columns  

5 

54 

Longitudinal  section  of  solid  

4 

51 

47 

Long-ton  table  

2 

10 

25 

Loss  or  gain  per  cent  

2 

6 

J.  Sec.  Page. 

Jib  crane,  Graphical  diagram  for      5      122 

K.  Sec.  Page. 

Keystone  octagonal  column 6       32 

Li.                    Sec.  Page. 

Larimer  column 6  31 

Lateral  bracing  of  roof  truss 6  131 

Latticed  angle  column 6  32 

Laws  of  motion,  Newton's 5  4 

League..- 2  11 

Least  common  denominator,  To 

find 1  26 

Lever 5  23 

"     arm  of  force 5  18 

"     Principle  of 5  24 

Like  numbers 1  1 


M.  Sec.  Page. 

Materials,  Elasticity  of 6  102 

Maximum  shear 5  72 

Means  (Proportion) 2  47 

Measure,  Definition  of 2  8 

"          Dry..... 2  10 

"         Linear 2  8 

"          of  angles  or  arcs 2  11 

"           "  money 2  11 

"           "  time 2  11 

"          Square 2  9 

"          Surveyors'  linear 2  8 

"                    "           square.....  2  9 

Measures,  Classification  of 2  8 

"          Cubic 2  9 

"          Liquid 2  10 

"          Miscellaneous 2  11 

"          of  capacity 2  10 

"           "  extension  2  8 

"           "  weight 2  9 

"          Standard  units  of 2  8 

Mechanics,  Definition  of 5  3 

"           Elements  of 5  3 

Mensuration,  Definition  of 4  23 

of  plane  surfaces. .  4  26 

"              "solids 4  44 

Meter 2  11 

Minuend 1  9 

Minus 1  9 

Mixed  number..  .  1  23 


INDEX.                                                                      XV11 

Sec. 

Page. 

Sec. 

/'«IAV. 

Mixed  number,  To  reduce,  to  im- 

Notation   

1 

1 

proper  fraction  

1 

25 

"         Arabic  

1 

2 

Moduli  of  elasticity  :  Table  

G 

148 

Number  

1 

1 

Modulus  of  elasticity  

0 

103 

"         Abstract  

1 

1 

"          "  rupture  

5 

42 

"         Concrete  

1 

1 

"           "  section  

5 

84 

"         Denominate  

2 

7 

Moment,  Bending  

5 

74 

"         Mixed  

1 

23 

"         of  a  force  

5 

17 

"         Prime  

1 

20 

"          "  inertia  

0 

8 

"         Unit  of  

1 

1 

"           "        "       determined 

Numbers,  Like  

1 

1 

by     graph- 

"        Reading  

1 

3 

ical     meth- 

"         Unlike  

1 

1 

ods  

6 

11 

Numeration  

1 

1 

"          "        "       Rules    and 

"            

1 

22 

formulas 

for  

6 

10 

(). 

Sec. 

Page. 

"           "  resistance  and  bend- 

Oblique triangle  

4 

8 

ing  moment,  Rela- 

Obtuse angle  

4 

3 

tion  between  

6 

14 

Octagon  

4 

G 

"          Resisting  

5 

75 

Origin  of  moments  

5 

18 

"                   "         

5 

84 

Osborn's  code  for  rivets  

G 

41 

"         Resultant  

5 

20 

Moments,  Algebraic  sum  of  

5 

20 

P. 

Sec. 

PaKe. 

"         Center  or  origin  of.  ... 

5 

18 

Packing  piece  for  plate  girder..  . 

G 

57 

"          due  to  loads,  Effect  of 

5 

78 

Panel  points  

5 

113 

"         Equilibrium  of  

5 

19 

Parallel  lines  

4 

1 

"         how  expressed  

5 

19 

Parallelogram  

4 

28 

"         Positive  and  negative 

5 

20 

"               Area  of  

4 

2!) 

Money,  Measure  of  

3 

11 

of  forces  

5 

G 

"        U.  S  

2 

11 

Parallelepiped  

4 

44 

Motion  

5 

3 

Parenthesis  

1 

49 

"        Newton's  laws  of  

5 

4 

"           

3 

2 

Multiplicand  

1 

11 

Pentagon  

4 

G 

Multiplication  

1 

11 

Per  cent..  Sign  of  

2 

1 

"              of  decimals  

1 

40 

Percentage  

2 

1 

"                "denominate 

"          

2 

2 

numbers  

2 

19 

"          Meaning  of  

2 

1 

"  fractions  

1 

31 

Perimeter,  Definition  of  

4 

6 

"              Rule  for  

1 

14 

"          of  ellipse  

4 

38 

"               Sign  of  

1 

11 

Permanent  set  

G 

102 

"              table  

1 

12 

Perpendicular  

4 

1 

Multiplier  

1 

11 

Phoenix  column  

G 

32 

Pin    subjected    to  bending 

X. 

Sec. 

Page. 

stresses  

6 

50 

Naught  ;  

1 

2 

Pins  and  eyes  of  trusses  

G 

133 

5 

20 

"    Resultant  moment  of 

5 

72 

stresses  on  

G 

52 

Neutral  axis  

5 

76 

"    Strength  of  

G 

43 

"          "     

6 

1 

"    Table  of  resisting  moments 

"          "    determined     by     a 

of  

6 

52 

graphical  method 

6 

5 

"    Table  of  resisting  moments 

"          "    determined     by     a 

of  

G 

149 

principle    of    mo- 

Plane figure,  Area  of  

4 

39 

ments  

0 

2 

"      figures  

4 

G 

6 

1 

"           "      Center  of  gravity  of 

5 

2G 

Newton's  laws  of  motion  

5 

4 

"     section  of  solid  

4 

51 

XV111 


INDEX. 


V,r. 

Page. 

Sec. 

Page. 

Plane  surface  

4 

6 

Pyramid,  Altitude  of  

4 

47 

"      surfaces,  Mensuration  of.. 

4 

26 

"          Frustum  of  

4 

49 

Plate  girder  

8 

57 

"          Right  

4 

47 

"           "        Depth  of  

6 

00 

"           Slant  height  of  

4 

47 

"           "       design,  Practical 

"           Surface  of  

4 

47 

problems  in  

6 

97 

"          Volume  of  

4 

48 

"           "       Flanges  of  

6 

67 

"           "       General    construc- 

Q- 

Sec. 

Page. 

tion  of  

6 

57 

4 

17 

"           "        Stiffness  of  

6 

57 

Quadrilateral  

4 

6 

Plate  girders,  Principles  of  de- 

"            Diagonal  of  

4 

29 

sign  of  

8 

59 

Quantity,  Meaning  of  term  

3 

1 

"            "          Usual  sections  of 

6 

58 

Quotient  

1 

16 

Plus  

1 

4 

Point  

4 

1 

R. 

Sec. 

Page. 

"     of  contact  

4 

19 

Radical  sign  

2 

25 

"       "  intersection  

4 

2 

"         "     

3 

2 

"       "  tangency  

4 

19 

Radii  of  gyration  for  two  angles 

6 

145 

Polonceau  stress  

8 

110 

Radius  of  circle  

4 

16 

Polygon  

4 

6 

"       "  gyration  

5 

57 

"        Area  of  

4 

30 

"       "          "         

6 

14 

"        Equiangular  

4 

7 

Rankine-Gordon  formula  

3 

10 

"        Equilateral  

4 

6- 

Rate  (per  cent.)  

2 

2 

"       Inscribed    and    circum- 

Ratio   

2 

42 

scribed  

4 

21 

•'      Direct.  

2 

43 

"       of  forces  

5 

11 

"      Inverse  

2 

43 

"        Regular  

4 

7 

"      Operations  upon  

2 

45 

"        Sum  of  angles  of  

4 

7 

"      Reciprocal  

2 

43 

Positive  and  negative  moments. 

6 

20 

"      Terms  of  

2 

43 

"         shear  

6 

72 

"      To  invert  

2 

44 

Power  of  a  number  

9 

22 

"      Value  of  

j> 

43 

Prime  factor  

1 

20 

Reactions  

5 

62 

"      number  t  

1 

20 

"         Relation  between  

5 

03 

Primes  and  subscripts  

8 

5 

Reciprocal  ratio  

2 

43 

Principles  of  successful  design.. 

5 

2 

Rectangle  

4 

28 

Prism,  Area  of  surface  of  

4 

45 

Reduction  of  decimals  

1 

46 

"       Definition  of  

4 

44 

"          "  denominate    num- 

Right   

4 

45 

bers  

2 

12 

"       Volume  of  

4 

46 

"          "  fractions  

1 

23 

Prismoid,  Definition  of  

4 

51 

Regular  polygon  

4 

7 

"         Volume  of  

4 

51 

Relative  and  local  values  of  fig- 

Prismoidal formula  

4 

51 

ures  

1 

2 

Product  

1 

11 

Remainder  

1 

9 

Proper  fractions  

1 

23 

Resisting  inches  

5 

84 

Proportion  

* 

46 

"         moment..  

5 

75 

"         Inverse  

8 

47 

"                "        

5 

84 

"              "        

a 

50 

"                "        

6 

14 

"         Compound  

» 

55 

"         momentsof  pins,  Table 

"         Direct  

9 

47 

of  

6 

52 

"         how  read  

9 

46 

"         momentsof  pins,  Table 

"         how  written  

» 

46 

of  

6 

149 

"         Operations  in  

9 

48 

Resultant  moment  of  several 

"         Powers  and  roots  in.. 

9 

52 

stresses  on  pin  

6 

52 

9 

47 

"          moments  

5 

30 

"         Simple  

9 

55 

"          of  forces  

5 

7 

Pyramid  

4 

47 

"          "        "     

5 

14 

INDEX. 


XIX 


Sec.  Page. 

Resultant  of  several  forces 5  10 

Rhomboid 4  28 

Rhombus 4  28 

Right  angle 4  3 

Right-angled  triangle 4  8 

Right  pyramid  or  cone 4  47 

"      section  of  a  solid 4  51 

"      triangle.  Application  of,  to 

roofs 4  12 

Rivet  signs 6  41 

"     spacing  in  girder.  Effect  of 

vertical  stress  upon 6  92 

"     spacing  in  girders 6  90 

"            "          "  the  flange  plates  6  % 
Rivets  and  end  angles  or  stiffen- 

ers  over  abutment  6  90 

"         "    pins,  Strength  of 6  43 

"         "    rivet   spacing  in  col- 
umns   6  33 

"      Bearing  value  of 6  45 

"      connecting  flange  angles 

with  web 6  90 

"       in  double  shear 6  45 

"        "  ordinary  bearing 6  45 

"        "  single  shear 6  45 

"        "  stiffeners      between 

abutments 6  90 

"        "  web  bearing 6  45 

"      spaced  according  to  bend- 
ing moment 6  93 

"      spaced  according  to  direct 

vertical  shear 6  96 

Riveted  joints,  Failure  of 6  44 

Roof  truss,  Composite  members 

of 6  132 

"         "      Composite     pin-con- 
nected. Design  of..  6  118 

"         "      Diagram  for 5  121 

"         "      Factors  of  safety  for  6  132 
"         "      General     notes     re- 
garding design  of. .  6  131 
*'•         "      Lateral  bracing  of. ..  6  131 
"         "       Structural-steel,  De- 
sign of 6  124 

"         "      Tension  members  of  6  132 

"         "      with  40- foot  span 5  123 

Roof  trusses 6  110 

"  "      Details  and  design 

of 6  134 

"           "       Pins  and  eyes  of 6  133 

Rolled  shapes.  Conventional 

signs  for 6  42 

Root,  Cube 2  25 

"     2  32 

"     Index  of! 2  25 

"     of  a  number. . .  2  23 


Sec.  Page. 

Root,  Square -  25 

Roots  of  fractions 2  40 

"      other     than     square    and 

cube 2  41 

Rule  for  addition 1  8 

"       "    cancelation 1  21 

"       "    compound-proportion..  2  56 

"       "    cube  root 2  38 

"       "    division 1  18 

"       "    multiplication 1  14 

"       "    square  root 2  30 

"       "    subtraction 1  10 

"      of  three 2  47 

Rules  for  percentage 2  2 

Rupture,  Modulus  of 5  42 

S.  Sec.  Page. 

Safety,  Factor  of 5  40 

Sandwiched  girder 6  100 

Scalene  triangle 4  8 

Scales,  Use  of 4  24 

Score 2  11 

Section  modulus 5  84 

"  "         of  channel,  Ap- 

proximate 

rule  for 5  87 

"          "  I  beam,   Ap- 
proximate 

rule  for 5  87 

Sections  of  a  solid 4  51 

"         "  structural    material, 

Elements  of 6  135 

"         used     in    plate -girder 

construction 6  58 

Sector  of  circle 4  17 

"        "      "      Area  of 4  36 

Segment  of  circle 4  16 

"           "      "      Area  of 4  37 

"  "      "       Formulas    for 

radius.chord, 

and  height  of  4  33 

Semicircle 4  18 

Semicircumference 4  18 

Separators  for  I  beams 5  94 

Shear 5  70 

"      and  bending  moment,  Re- 

lation  between 5  79 

"      in  a  simple  beam 5  70 

'.'      Maximum 5  72 

"      Positive  and  negative 5  72 

"      Vertical 5  72 

Shearing  stress 5  3G 

"           stresses  in  web-plate..  6  60 

Short  columns 5  53 

Sign  for  dollars 1  48 

"     of  addition 1  4 


XX 


INDEX. 


Sec.  Page. 

Sign  of  division 1  16 

"      "  equality 1  4 

"     "  multiplication 1  11 

"     "  percent 2  1 

"     "  subtraction 1  9 

"     Radical 2  25 

"       3  2 

Similar  figures 4  55 

"  "       Relations  of  areas 

and  volumes  of  4  55 
"             "       Relations  of  areas 

and  volumes  of  4  56 

"        triangles 4  13 

Simple  beam 5  62 

"        denominate  number 2  7 

"       proportion 2  55 

Slant  height  of  pyramid  or  cone  4  47 

Snow  and  wind  loads 5  33 

Solid,  Convex  surface  of 4  44 

"      Cubical  contents  of 4  44 

"      Definition  of 4  44 

"      Entire  surface  of 4  44 

"      Faces  and  edges  of 4  44 

"      Sections  of 4  51 

"      Volume  of 4  44 

Solids,  Mensuration  of 4  44 

Spacing  stiffeners,  Practical  rule 

for 6  65 

Span  and  depth  of  beam,  Rela- 
tion between 5  94 

"      of  arch 4  20 

"       "  simple  beam 5  62 

Sphere 4  53 

"       Area  of  surface  of 4  53 

"       Diameter  of,  determined 

from  volume. ..   4  54 

"       Volume  of 4  53 

Splices  and  connections  for  col- 
umns   6  34 

Square 4  28 

"       foot,  Definition  of 4  26  . 

"       inch,  Definition  of 4  26 

"       measure 2  9 

"        Surveyors' 2  9 

"       root 2  25 

"           "  Proof  of 2  30 

"          "  Rule  for 2  30 

"          "  Short  method  for 2  31 

Stable  equilibrium 5  16 

Standard  units  of  various  meas- 
ures    2  8 

Statics 5  3 

Steel  beams 5  86 

"          "      Strength  of 5  87 

"          "      Varieties  of. v 5  86 

"      columns,  Forms  of 6  26 


Sec.  Page. 

Steel  columns,  Formulas  for 6  18 

"             "          Types  of 6  16 

Stiffeners  for  plate  girder 6  57 

"          of  girder,  Distribution 

of 6  63 

"          Practical  rule  for  spa- 
cing    6  65 

Stiffness,  Definition  of 6  104 

"          of  plate  girder 6  47 

Stone  beams,  Strength  of 5  99 

Straight  line 4  1 

Strain,  Definition  of 5  39 

" 5  41 

Unit 5  41 

Strength  of  beams 5  84 

"           "  building  materials. .  5  42 
"           "  foundation    materi- 
als   5  48 

"           "  rivets  and  pins 6  43 

"          "  steel  beams 5  87 

"         Ultimate 5  42 

Stress  and  frame  diagrams,  Or- 
der of  letters  in 5  115 

"        Compressive 5  39 

"       Definition  of 5  39 

"        diagram 5  112 

"        Intensity  of 5  41 

"        Shearing 5  39 

"       Tensile 5  39 

"       Transverse  or  bending. . .  5  40 

"       Twisting 5  40 

"        Unit 5  41 

Stresses  and  flange  of  girder 6  68 

"        Bending 5  74 

"        Graphical     method     of 

computing 5  105 

"        in  beams 5  69 

"         "      "      with  one  strut..  5  102 
"         "      "         "     two  struts  5  103 
"         "  Fink  truss,  Determin- 
ing   6  110 

Subscripts  and  primes 3  5 

Subtraction 1  9 

"            of  decimals 1  39 

"             "  denominate  num- 
bers   2  17 

"             "  fractions 1  29 

"             Rule  for 1  10 

"             Sign  of 1  9 

Subtrahend 1  9 

Surface,  Definition  of 4  6 

"          of  prism  or  cylinder. ..  4  45 

"           "  solid 4  44 

"           "  sphere,  Area  of 4  53 

"          Plane 4  6 

Surfaces,  Mensuration  of 4  26 


INDEX. 


XXI 


Sec.  Page. 

2       s 

2          9 
49 


Surveyors'  linear  measure 2 

"  square  measure 2 

Symbols  of  aggregation 1 

"         "  "  Order  of 

prece- 
denc  e 

in 

Symmetrical  figures 4 

Symmetry,  Axis  of 4 

T. 

Tangency,  Point  of 4 

Tangent  to  circle 4 

Tees 

Tensile  stress 

Tension  members  of  roof  truss. 

Terms  of  a  fraction 

"        "  "  ratio 

Time,  Measure  of 2 

Torsion 

Transverse  stress 5 

Trapezium 4 

Trapezoid 4 

"          Area  of 

Triangle 4 

"         Area  of 

"         Equilateral 4 

"         General  properties  of.. 

"         Isosceles  

"         Oblique 4 

"         offerees 

"         Right-angled 4 

"         Scalene 

Triangles,  Equality  of 4 

"  Similar 4 

Troy  weight 2 

Truss,Designing  the  members  of 

"     for  church  roof 5 

Trussed  beams 

Trusses,  Members  in,  subjected 
to  transverse  and  di- 
rect stresses 

"         Members  in,  subjected 
to  transverse  stresses 

"         Pins  and  eyes  of 6 

"         Roof 6 

Twisting  stress 5 


TJ.  Sec.  Page. 

Ultimate  elongation 5       42 

"         strength 5       42 

Unit 1         1 

"    of  a  number....  1         1 


Sec.  Page. 

Unit  square 4  20 

"    strain 5  41 

"    stress 5  41 

U.  S.  money 2  11 

Unlike  numbers 1  1 


1 

49 

4 

54 

V. 

Sec. 

Page. 

4 

54 

Value  of  a  fraction  

1 

23 

"       "  "  ratio  

2 

43 

tec. 

Page. 

Value  of  rivets,  Table  of.,.. 

6 

49 

4 

19 

C 

13(i 

4 

19 

Vary,  Meaning  of  the  term  

2 

4% 

5 

86 

Vertex  of  angle  

4 

2 

5 

39 

Vertical  line  

4 

2 

6 

132 

"          shear  

5 

72 

1 

23 

"          stress  in  flange,  Effect  of 

6 

92 

2 

43 

Vinculum  

1 

49 

2 

11 

"         

3 

2 

5 

40 

Volume  of  frustum  of  pyramid 

5 

40 

or  cone  

4 

50 

4 

28 

"          "  prism  or  cylinder  

4 

40 

4 

28 

"         "  prismoid  

4 

51 

4 

29 

"          "  pyramid  or  cone  

4 

48 

4 

6 

"         "  solid  

4 

44 

4 

27 

"         "  sphere  

4 

53 

4 

8 

"         "  wedge  -.  

4 

50 

4 

9 

4 

8 

W. 

Sec. 

Page. 

4 

8 

Web-plate,  Buckling  of  

6 

W 

5 

9 

"        "         of  girder  

6 

57 

4 

8 

"        "          "      "        Shearing 

4 

8 

stresses 

4 

13 

in  

6 

60 

4 

13 

"        "         "      "        Thickness 

2 

10 

of  

6 

61 

5 

139 

Wedge,  Definition  of  

4 

50 

5 

126 

"        Volume  of  

4 

50 

5 

101 

Weight,  Avoirdupois  

2 

9 

"        Measures  of  

2 

9 

"        of  building  materials.  .. 

5 

27 

6 

133 

Wind  and  snow  loads  

5 

33 

"     diagram  

5 

129 

e 

133 

Wood  columns,  Formula  for.... 

5 

54 

6 

133 

Wooden  truss,  80-foot  span  

5 

123 

6 

110 

Working  drawings,  how  dimen- 

5 

40 

sioned 

4 

24 

"                  "           how  made,  . 

4 

21 

Z.  Sic.  Page. 

Z-bar  column I       31 

2  bars,  Properties  of  :  Table    ...      6      141 
Zero...  1         2 


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